HDU 1839 Delay Constrained Maximum Capacity Path 最短路+二分

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题目来源：HDU 1839 Delay Constrained Maximum Capacity Path

题意：给你一张无向图要从1到n运送东西每条路都有容量限制和经过的时间选择一条路径该路径最多能运的数量等于容量最少的那条边并且总时间不能超过T

思路：和上一题一样二分容量然后做最短路判断是否最短时间小于等于T

#include <cstdio>#include <algorithm>#include <queue>#include <vector>using namespace std;const int maxn = 50010;struct edge{int u, v, h, w;};struct HeapNode{int u, d;bool operator < (const HeapNode& rhs)const{return d > rhs.d;}};vector <edge> G[maxn];int dis[maxn];bool vis[maxn];int n, m, t;void Dijkstra(int h){for(int i = 0; i <= n; i++)dis[i] = 999999999;//memset(dis, 0x7f, sizeof(dis));dis[1] = 0;memset(vis, false, sizeof(vis));priority_queue <HeapNode> Q;Q.push((HeapNode){1, 0});while(!Q.empty()){HeapNode x = Q.top();Q.pop();int u = x.u;if(vis[u])continue;vis[u] = true;for(int i = 0; i < G[u].size(); i++){ edge e = G[u][i];if(e.h < h)continue;int v = e.v;if(dis[v] > x.d + e.w){dis[v] = x.d + e.w;Q.push((HeapNode){v, dis[v]});}}}}int main(){int T;scanf("%d", &T);while(T--){scanf("%d %d %d", &n, &m, &t);for(int i = 0; i <= n; i++)G[i].clear();for(int i = 0; i < m; i++){int u, v, h, w;scanf("%d %d %d %d", &u, &v, &h, &w);G[u].push_back((edge){u, v, h, w});G[v].push_back((edge){v, u, h, w});}int l = 0, r = 2000000000, ans = -1;while(l <= r){int mid = (l + r) >> 1;Dijkstra(mid);if(dis[n] <= t){ans = mid;l = mid + 1;}else{r = mid - 1;}}printf("%d\n", ans);}return 0;}

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