Contest Results

(Analysis by Benjamin Qi, Danny Mittal)

If the graph is bipartite, the answer is $N-1$. Otherwise, let

$$g(i)=(dist_{even}(i),dist_{odd}(i))$$

for each vertex $1\le i\le N$, the same as "Sum of Distances" from the January contest. Also define

$$h(i)=(\min(dist_{even}(i),dist_{odd}(i)),\max(dist_{even}(i),dist_{odd}(i)))=(a_i,b_i).$$

For convenience, define $s(a, b)$ to be the set of all nodes $i$ with $h(i) = (a, b)$.

A graph having $f_G$ equivalent to that of the given graph must have edges adjacent to vertex $i$ satisfying at least one of the following conditions for each $2\le i\le N$ (the condition is slightly different for vertex $1$ since $a_1=0$).

$i$ is adjacent to a vertex in $s(a_i-1,b_i-1)$. Call this an edge of type A.
If $a_i+1<b_i$, then $i$ is adjacent to a vertex in $s(a_i-1,b_i+1)$ and a vertex in $s(a_i+1,b_i-1)$. Call these edges of type B.
If $a_i+1=b_i$, then $i$ is adjacent to a vertex in $s(a_i-1,b_i+1)$ and a vertex in $s(a_i,b_i)$ (possibly $i$ itself). Call an edge from $i$ to a vertex in $s(a_i, b_i)$ an edge of type C.

Edges that are not type A, B, or C edges cannot exist in the graph.

We can look at each layer (vertices with a fixed $a_i+b_i$) independently, as edges of type A are only relevant to the vertex in the higher layer, and edges of type B and C are between vertices in the same layer. For a given layer, let's satisfy the constraints in increasing order of $a_i$.

We'll describe two ways of doing this, of which the second approach is sufficient for full credit.

Suppose that we are currently deciding which edges to construct involving $s_{a,b}$. For simplicity, we'll only deal with the case that $|s_{a-1,b-1}|>0$ and $|s_{a+1,b-1}|>0$ (for the other cases, see the code for details). Our goal is to satisfy one of the first two conditions for each vertex $v$.

Approach 1: Define:

$j$ as the number of type B edges from $s_{a,b}$ to $s_{a-1,b+1}$
$k$ as the number of type B edges from $s_{a,b}$ to $s_{a+1,b-1}$

Then we'll need to add $\max(|s_{a,b}|-\min(j,k),0)$ additional edges of type A.

These observations are sufficient for an $\mathcal{O}(N^2)$ DP. Store $dp_{a,b}[j]$ for each $0\le j\le \max(|s_{a-1,b+1}|,|s_{a,b}|)$ and transition to $dp_{a+1,b-1}[k]$ for each $0\le k\le \max(|s_{a,b}|,|s_{a+1,b-1}|)$. See my $\texttt{solve_between}$ function below for details:

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python!
 
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;
 
#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;
 
// pairs
#define mp make_pair
#define f first
#define s second
 
// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back 
#define pf push_front
#define rtn return
 
#define lb lower_bound
#define ub upper_bound 
tcT> int lwb(V<T>& a, const T& b) { return int(lb(all(a),b)-bg(a)); }
 
// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for (auto& a: x)
 
const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;
 
// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { // assert(x >= 0); // make C++11 compatible until USACO updates ...
	return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }
 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down
 
tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }
 
tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(it); } // element that doesn't exist from (multi)set
 
// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);
 
tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }
 
tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(V<T>& x) { each(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { each(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }
 
// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	// #ifdef LOCAL
	// 	return b ? "true" : "false"; 
	// #else 
	return ts((int)b);
	// #endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(V<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;
 
	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}
 
// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }
 
// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif
 
void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void setIO(str s = "") {
	unsyncIO(); setPrec();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) setIn(s+".in"), setOut(s+".out"); // for USACO
}
 
#define ints(...); int __VA_ARGS__; re(__VA_ARGS__);
 
 
int N,M;
vi adj[MX];
 
V<AR<int,2>> gen_dist() {
	V<AR<int,2>> dist(N,{MOD,MOD});
	queue<pi> q;
	auto ad = [&](int a, int b) {
		if (dist[a][b%2] != MOD) return;
		dist[a][b%2] = b; q.push({a,b});
	};
	ad(0,0);
	while (sz(q)) {
		pi p = q.ft; q.pop();
		each(t,adj[p.f]) ad(t,p.s+1);
	}
	return dist;
}
 
int ans = 0;
set<pi> distinct;
 
int div2(int x) { return (x+1)/2; }
 
void solve_between(vi nums, vb exists, bool special) {
	vi dp{0};
	int res = MOD;
	F0R(i,sz(nums)) {
		if (i == sz(nums)-1) {
			if (special) {
				F0R(j,sz(dp)) F0R(k,nums[i]+1) {
					int need_one = max(nums[i]-min(j,2*k),0);
					if (!exists[i] && need_one) continue;
					ckmin(res,dp[j]+need_one+k);
				}
			} else {
				assert(exists[i]);
				each(t,dp) ckmin(res,t+nums[i]);
			}
		} else {
			vi DP(max(nums[i],nums[i+1])+1,MOD);
			F0R(j,sz(dp)) F0R(k,max(nums[i],nums[i+1])+1) {
				int need_one = max(nums[i]-min(j,k),0);
				if (!exists[i] && need_one) continue;
				ckmin(DP[k],dp[j]+need_one+k);
			}
			swap(dp,DP);
		}
	}
	ans += res;
}
 
void solve_sum(int sum, vpi v) {
	dbg("SOLVE SUM",sum,v);
	assert(sz(v));
	if (v[0].f == 0) {
		F0R(i,sz(v)-1) ans += max(v[i].s,v[i+1].s);
		ans += div2(v.bk.s);
		return;
	}
	for (int i = 0; i < sz(v); ++i) {
		vi nums{v[i].s};
		vb exists{distinct.count({v[i].f-1,sum-v[i].f-1})};
		while (i+1 < sz(v) && v[i+1].f == v[i].f+1) {
			++i; nums.pb(v[i].s);
			exists.pb(distinct.count({v[i].f-1,sum-v[i].f-1}));
		}
		bool special = 0;
		if (2*v[i].f+1 == sum) special = 1;
		solve_between(nums,exists,special);
	}
}
 
void go() {
	auto a = gen_dist();
	if (a[0][1] == MOD) {
		ps(N-1);
		return;
	}
	map<int,map<int,int>> cnt;
	each(t,a) {
		pi p{t[0],t[1]};
		if (p.f > p.s) swap(p.f,p.s);
		distinct.ins(p);
		++cnt[p.f+p.s][p.f];
	}
	each(t,cnt) solve_sum(t.f,vpi(all(t.s)));
	ps(ans);
}
 
int main() {
	setIO(); 
	int T; re(T);
	F0R(_,T) {
		re(N,M);
		F0R(i,N) adj[i].clear();
		ans = 0; distinct.clear();
		F0R(i,M) {
			int a,b; re(a,b); --a,--b;
			adj[a].pb(b), adj[b].pb(a);
		}
		go();
	}
}
/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/

Approach 2: For full credit, we can assign edges greedily.

Suppose that we have already added $\texttt{prev}$ edges from $s_{a,b}$ to $s_{a+1,b-1}$ in order to satisfy the conditions for vertices $w$. Let's assign each of these edges to different vertices $v$ if possible, meaning that we have at least $x=\min(\texttt{prev},|s_{a,b}|)$ vertices $v$ with edges to $(a-1,b+1)$.

If we cannot add edges of type A (no vertex exists with $(a-1,b-1)$), then we must satisfy the second condition for each vertex $v$. We'll need to add $|s_{a,b}|-x$ additional edges of type B to $(a-1,b+1)$ and $|s_{a,b}|$ edges to $(a+1,b-1)$.
Otherwise, we need to choose whether to satisfy the first or the second condition for each vertex.
- For each of the $x$ vertices that already have an edge of type B to $(a-1,b+1)$, we can satisfy the first condition by adding an edge of type A to $(a-1,b-1)$ or satisfy the second condition by adding an edge of type B to $(a+1,b-1)$. It's always at least as good to do the latter. Both options add a single edge, and in the case of a tie it's always better to increase the number of edges between $(a,b)\Leftrightarrow (a+1,b-1)$.
- For each of the $|s_{a,b}|-x$ vertices without an edge of type B to $(a-1,b+1)$, we can satisfy the first condition by adding an edge of type A to $(a-1,b-1)$ or add two edges of type B, one to $(a-1,b+1)$ and the other to $(a+1,b-1)$. It's always at least as good to do the former, since it results in the addition of only a single edge (we can always add an additional edge between $(a,b)\Leftrightarrow (a+1,b-1)$ later on).

Danny's code (which doesn't explicitly group vertices by $a+b$):

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class MinimizingEdgesActuallyCorrect {
 
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tokenizer = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        int m = Integer.parseInt(tokenizer.nextToken());
        List<Integer>[] adj = new List[(2 * n) + 1];
        for (int a = 1; a <= 2 * n; a++) {
            adj[a] = new ArrayList<>();
        }
        for (int j = 1; j <= m; j++) {
            tokenizer = new StringTokenizer(in.readLine());
            int a = Integer.parseInt(tokenizer.nextToken());
            int b = Integer.parseInt(tokenizer.nextToken());
            adj[a].add(b + n);
            adj[b + n].add(a);
            adj[a + n].add(b);
            adj[b].add(a + n);
        }
        int[] dist = new int[(2 * n) + 1];
        Arrays.fill(dist, -1);
        dist[1] = 0;
        LinkedList<Integer> q = new LinkedList<>();
        q.add(1);
        while (!q.isEmpty()) {
            int a = q.remove();
            for (int b : adj[a]) {
                if (dist[b] == -1) {
                    dist[b] = dist[a] + 1;
                    q.add(b);
                }
            }
        }
        int answer = 0;
        if (dist[n + 1] == -1) {
            answer = n - 1;
        } else {
            TreeMap<Pair, Integer> freq = new TreeMap<>();
            TreeMap<Pair, List<Integer>> buckets = new TreeMap<>();
            for (int a = 1; a <= n; a++) {
                freq.merge(new Pair(Math.min(dist[a], dist[n + a]), Math.max(dist[a], dist[n + a])), 1, Integer::sum);
                buckets.computeIfAbsent(new Pair(Math.min(dist[a], dist[n + a]), Math.max(dist[a], dist[n + a])), __ -> new ArrayList<>()).add(a);
            }
            TreeMap<Pair, Integer> edgeAmt = new TreeMap<>();
            for (Map.Entry<Pair, Integer> entry : freq.entrySet()) {
                Pair p = entry.getKey();
                int f = entry.getValue();
                int prev = edgeAmt.getOrDefault(new Pair(p.first - 1, p.second + 1), 0);
                if (p.second == p.first + 1) {
                    if (p.first == 0) {
                        answer += (f + 1) / 2;
                    } else if (freq.containsKey(new Pair(p.first - 1, p.second - 1))) {
                        answer += Math.max((f - prev) + ((prev + 1) / 2), (f + 1) / 2);
                    } else {
                        if (prev < f) {
                            answer += f - prev;
                        }
                        answer += (f + 1) / 2;
                    }
                } else {
                    answer += f;
                    if (p.first == 0) {
                        edgeAmt.put(p, f);
                    } else if (freq.containsKey(new Pair(p.first - 1, p.second - 1))) {
                        edgeAmt.put(p, Math.min(f, prev));
                    } else {
                        if (prev < f) {
                            answer += f - prev;
                        }
                        edgeAmt.put(p, f);
                    }
                }
            }
        }
        System.out.println(answer);
    }
 
    static class Pair implements Comparable<Pair> {
        final int first;
        final int second;
 
        Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
 
        @Override
        public int compareTo(Pair other) {
            if (first != other.first) {
                return first - other.first;
            } else {
                return second - other.second;
            }
        }
    }
}

My code (which constructs a solution):

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python!
 
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;
 
#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;
 
// pairs
#define mp make_pair
#define f first
#define s second
 
// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back 
#define pf push_front
#define rtn return
 
#define lb lower_bound
#define ub upper_bound 
tcT> int lwb(V<T>& a, const T& b) { return int(lb(all(a),b)-bg(a)); }
 
// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for (auto& a: x)
 
const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;
 
// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { // assert(x >= 0); // make C++11 compatible until USACO updates ...
	return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }
 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down
 
tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }
 
tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(it); } // element that doesn't exist from (multi)set
 
// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);
 
tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }
 
tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(V<T>& x) { each(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { each(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }
 
// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	// #ifdef LOCAL
	// 	return b ? "true" : "false"; 
	// #else 
	return ts((int)b);
	// #endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(V<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;
 
	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}
 
// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }
 
// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif
 
void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void setIO(str s = "") {
	unsyncIO(); setPrec();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) setIn(s+".in"), setOut(s+".out"); // for USACO
}
 
#define ints(...); int __VA_ARGS__; re(__VA_ARGS__);
 
 
int N,M;
vi adj[MX];
 
V<AR<int,2>> gen_dist() {
	V<AR<int,2>> dist(N,{MOD,MOD});
	queue<pi> q;
	auto ad = [&](int a, int b) {
		if (dist[a][b%2] != MOD) return;
		dist[a][b%2] = b; q.push({a,b});
	};
	ad(0,0);
	while (sz(q)) {
		pi p = q.ft; q.pop();
		each(t,adj[p.f]) ad(t,p.s+1);
	}
	return dist;
}
 
vpi ans_ed;
map<pi,int> distinct;
vpi group;
 
int div2(int x) { return (x+1)/2; }
 
void satisfy(vi a, vi b) {
	F0R(i,max(sz(a),sz(b))) ans_ed.pb({a[min(i,sz(a)-1)],b[min(i,sz(b)-1)]});
}
 
void satisfy_self(vi a) {
	for (int i = 0; i < sz(a); i += 2) {
		if (i == sz(a)-1) ans_ed.pb({a[i],a[i]});
		else ans_ed.pb({a[i],a[i+1]});
	}
}
 
void satisfy_lower_left(vi v) {
	each(t,v) {
		pi p = group[t]; --p.f,--p.s;
		assert(distinct.count(p));
		ans_ed.pb({t,distinct[p]});
	}
}
 
void satisfy_upper_left(vi v) {
	each(t,v) {
		pi p = group[t]; --p.f,++p.s;
		assert(distinct.count(p));
		ans_ed.pb({t,distinct[p]});
	}
}
 
void solve_between(V<vi> nums, vb exists, bool special) {
	vi bef;
	F0R(i,sz(nums)) {
		vi yes = nums[i], no; 
		while (sz(yes) > sz(bef)) {
			no.pb(yes.bk);
			yes.pop_back();
		}
		satisfy(bef,yes);
		if (i == sz(nums)-1) {
			if (special) {
				// ans += nums[i]-sz(bef);
				if (exists[i]) {
					satisfy_lower_left(no);
					// for each in yes: loop
					// for each in no: go to
					// ans += div2(bef);
					satisfy_self(yes);
				} else {
					satisfy_upper_left(no);
					satisfy_self(nums[i]);
				}
			} else {
				assert(exists[i]);
				satisfy_lower_left(nums[i]);
			}
		} else if (exists[i]) {
			bef = yes;
			satisfy_lower_left(no);
		} else {
			satisfy_upper_left(no);
			bef = nums[i];
		}
	}
}
 
void solve_sum(int sum, V<pair<int,vi>> v) {
	dbg("SOLVE SUM",sum,v);
	assert(sz(v));
	if (v[0].f == 0) {
		F0R(i,sz(v)-1) satisfy(v[i].s,v[i+1].s);
		satisfy_self(v.bk.s);
		return;
	}
	for (int i = 0; i < sz(v); ++i) {
		V<vi> nums{v[i].s};
		vb exists{distinct.count({v[i].f-1,sum-v[i].f-1})};
		while (i+1 < sz(v) && v[i+1].f == v[i].f+1) {
			++i; nums.pb(v[i].s);
			exists.pb(distinct.count({v[i].f-1,sum-v[i].f-1}));
		}
		bool special = 0;
		if (2*v[i].f+1 == sum) special = 1;
		solve_between(nums,exists,special);
	}
}
 
 
void solve() {
	auto a = gen_dist();
	// dbg(a);
	if (a[0][1] == MOD) {
		ps(N-1);
		return;
	}
	map<int,map<int,vi>> cnt;
	F0R(i,N) {
		AR<int,2> t = a[i];
		pi p{t[0],t[1]};
		if (p.f > p.s) swap(p.f,p.s);
		group.pb(p);
		distinct[p] = i;
		cnt[p.f+p.s][p.f].pb(i);
	}
	each(t,cnt) solve_sum(t.f,V<pair<int,vi>>(all(t.s)));
	F0R(i,N) adj[i].clear();
	each(t,ans_ed) adj[t.f].pb(t.s), adj[t.s].pb(t.f);
	assert(a == gen_dist());
	ps(sz(ans_ed));
	assert(sz(ans_ed) <= M);
}
 
int main() {
	setIO(); 
	int TC; re(TC);
	F0R(_,TC) {
		re(N,M);
		distinct.clear();
		F0R(i,N) adj[i].clear();
		ans_ed.clear(); group.clear();
		F0R(i,M) {
			int a,b; re(a,b); --a,--b;
			adj[a].pb(b), adj[b].pb(a);
		}
		solve();
	}
}