(Analysis by Jonathan Paulson - jonathanpaulson@gmail.com)

The first step is to see that we can break Bessie and Elsie's trip into two parts: the start until they meet and piggyback, and then piggybacking to the end (if they never piggyback, that's the same as meeting at the end and then piggybacking for 0 distance).

If we knew where they met, the problem would be easy. Suppose they meet at field $X$. Then from $X$, we can use a BFS to calculate the fastest paths to all the other fields. Then if it takes $D_i$ edges to travel from field $X$ to field $i$, the answer is $B*D_1 + E*D_2 + P*D_N$.

This already gives us an $O(N^2)$ solution; just try out each $X$. To speed it up to $O(N)$, notice that we always want the distances from fields $1$, $2$, and $N$. So if we precompute those distances, it takes $O(1)$ time to try out each $X$.

Here's my C++ code for that solution:

#include <cstdio> #include <vector> #include <queue> #define PB push_back using namespace std; typedef long long ll; vector<ll> D(ll st, const vector<vector<ll>>& E) { vector<ll> D(E.size(), -1); deque<ll> Q; D[st] = 0; Q.PB(st); while(!Q.empty()) { ll x = Q.front(); Q.pop_front(); for(ll y : E[x]) { if(D[y] == -1) { D[y] = D[x]+1; Q.PB(y); } } } return D; } int main() { ll b, e, p, n, m; scanf("%lld %lld %lld %lld %lld", &b, &e, &p, &n, &m); vector<vector<ll>> E(n, vector<ll>()); for(ll i=0; i<m; i++) { ll x, y; scanf("%lld %lld", &x, &y); x--; y--; E[x].PB(y); E[y].PB(x); } vector<ll> D0 = D(0, E); vector<ll> D1 = D(1, E); vector<ll> Dn = D(n-1, E); ll ans = ll(1000)*1000*1000*1000; for(ll meet=0; meet<n; meet++) { ans = min(ans, D0[meet]*b + D1[meet]*e + Dn[meet]*p); } printf("%lld\n", ans); }