二分+dijk算法
#include <iostream> #include <queue> #include <map> #include <set> #include <cmath> #include <cstring> #include <algorithm> #include <iomanip> #include <stack> #include <numeric> #include <ctime> #include <string> #include <bitset> #include <unordered_map> #include <unordered_set> using namespace std; using ll = long long; const ll N = 5e5 + 5, mod = 1e9 + 7, inf = 2e18; const double esp = 1e-8; int n, m, h; struct Node { ll v, w, z; }; vector<Node>g[N]; bool vis[N]; ll dis[N]; struct node { ll id, vlu; bool operator<(const node& u)const { return u.vlu < vlu; } }; bool dijk(int mid) { for (int i = 0; i <= n + 3; i++) { dis[i] = inf; vis[i] = false; } priority_queue<node>pq; pq.push({1, dis[1] = 0}); while (!pq.empty()) { node tem = pq.top(); pq.pop(); if (vis[tem.id])continue; vis[tem.id] = true; for (auto [y, w, z] : g[tem.id]) { if (dis[y] > dis[tem.id] + z && w >= mid) { pq.push({y, dis[y] = dis[tem.id] + z}); } } } return dis[n] <= h; } void solve() { cin >> n >> m >> h; while (m--) { ll u, v, w, z; cin >> u >> v >> w >> z; g[v].push_back({u, w, z}); g[u].push_back({v, w, z}); } ll l = 0, r = 1e9 + 5, ans = inf; while (l <= r) { ll mid = l + r >> 1; if (dijk(mid)) { l = mid + 1; ans = mid; } else { r = mid - 1; } } cout << (ans == inf ? -1 : ans); } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t = 1; //cin >> t; while (t--) { solve(); } return 0; }