(Analysis by Nick Wu)

To solve the first subtask, we can brute force all possible pairs of edges to add. This solution has complexity $\mathcal{O}(N^4(N+M))$.

To solve the second subtask, we need to be more disciplined in what edges we consider adding. We can start by computing the connected components of the barns - that is, maximal collections of barns where one can reach any barn from any other barn in the collection.

There are therefore two cases to consider here. We can either add an edge from the connected component containing barn 1 to the connected component containing barn $N$, or we pick an intermediate connected component, add one edge from it to the connected component containing barn $1$, and add another edge from it to the connected component containing barn $N$.

This observation on its own only reduces the number of pairs of edges to consider in the worst case by a constant factor. However, we now have two independent instances of the same subproblem to solve - given two connected components, what is the minimum cost needed to connect them with a single edge?

We can naively brute force this for all pairs of components we care about, for an $\mathcal{O}(N^2)$ algorithm.

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>

using namespace std;

void dfs(const vector<vector<int>>& edges, vector<int>& component, const int currv, const int id) {
for(int child: edges[currv]) {
if(component[child] != id) {
component[child] = id;
dfs(edges, component, child, id);
}
}
}

void solve() {
int n, m;
cin >> n >> m;
vector<vector<int>> edges(n);
for(int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
a--; b--;
edges[a].push_back(b);
edges[b].push_back(a);
}
vector<int> component(n);
iota(component.begin(), component.end(), 0);
for(int i = 0; i < n; i++) {
if(component[i] == i) {
dfs(edges, component, i, i);
}
}
if(component[0] == component[n-1]) {
cout << "0\n";
return;
}
vector<vector<int>> componentToVertices(n);
for(int i = 0; i < n; i++) {
componentToVertices[component[i]].push_back(i);
}
long long ans = 1e18;
vector<long long> srccost(n, 1e9);
vector<long long> dstcost(n, 1e9);
for(int i: componentToVertices[component[0]]) {
for(int j = 0; j < n; j++) {
srccost[component[j]] = min(srccost[component[j]], (long long)abs(i - j));
}
}
for(int i: componentToVertices[component[n-1]]) {
for(int j = 0; j < n; j++) {
dstcost[component[j]] = min(dstcost[component[j]], (long long)abs(i - j));
}
}
for(int i = 0; i < n; i++) ans = min(ans, srccost[i]*srccost[i] + dstcost[i]*dstcost[i]);
cout << ans << "\n";
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
for(int i = 0; i < t; i++) {
solve();
}
return 0;
}



To optimize this to $\mathcal{O}(M+N)$ time, we have to take advantage of the cost function.

For a given barn and a given connected component we want to connect it to, we only care about the smallest integer in the connected component greater than it, as well as the largest integer in the component less than it.

There are a couple ways to find these integers. One way would be to use binary search. It is also possible to do this in linear time by maintaining the pair of indices we consider and considering adding edges from vertex $i$ in increasing order of $i$. The pair of indices we consider will be nondecreasing, so we only consider a linear number of edges.

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>

using namespace std;

void dfs(const vector<vector<int>>& edges, vector<int>& component, const int currv, const int id) {
for(int child: edges[currv]) {
if(component[child] != id) {
component[child] = id;
dfs(edges, component, child, id);
}
}
}

void solve() {
int n, m;
cin >> n >> m;
vector<vector<int>> edges(n);
for(int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
a--; b--;
edges[a].push_back(b);
edges[b].push_back(a);
}
vector<int> component(n);
iota(component.begin(), component.end(), 0);
for(int i = 0; i < n; i++) {
if(component[i] == i) {
dfs(edges, component, i, i);
}
}
if(component[0] == component[n-1]) {
cout << "0\n";
return;
}
vector<vector<int>> componentToVertices(n);
for(int i = 0; i < n; i++) {
componentToVertices[component[i]].push_back(i);
}
long long ans = 1e18;
vector<long long> srccost(n, 1e9);
vector<long long> dstcost(n, 1e9);
int srcidx = 0;
int dstidx = 0;
for(int i = 0; i < n; i++) {
while(srcidx < componentToVertices[component[0]].size()) {
srccost[component[i]] = min(srccost[component[i]], (long long)abs(i - componentToVertices[component[0]][srcidx]));
if(componentToVertices[component[0]][srcidx] < i) srcidx++;
else break;
}
if(srcidx) srcidx--;
while(dstidx < componentToVertices[component[n-1]].size()) {
dstcost[component[i]] = min(dstcost[component[i]], (long long)abs(i - componentToVertices[component[n-1]][dstidx]));
if(componentToVertices[component[n-1]][dstidx] < i) dstidx++;
else break;
}
if(dstidx) dstidx--;
}
for(int i = 0; i < n; i++) ans = min(ans, srccost[i]*srccost[i] + dstcost[i]*dstcost[i]);
cout << ans << "\n";
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
for(int i = 0; i < t; i++) {
solve();
}
return 0;
}