HDU 2962 Trucking 最短路+二分

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题目来源 HDU 2962 Trucking

题意：给你一张无向图n个点m条边给出起点s终点e和最大承受的高度其中每条路都有限制的高度以及该条路的长度求从s到e最大可以通过的高度和在最大高度的前提下的最短路

思路：二分高度再求最短路无解特判一下

#include <cstdio>#include <algorithm>#include <queue>#include <vector>using namespace std;const int maxn = 1010;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;int s, e;void Dijkstra(int h){for(int i = 0; i <= n; i++)dis[i] = 999999999;//memset(dis, 0x7f, sizeof(dis));dis[s] = 0;memset(vis, false, sizeof(vis));priority_queue <HeapNode> Q;Q.push((HeapNode){s, 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 cas = 0;while(scanf("%d %d", &n, &m) && (n+m)){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);if(h == -1)h = 999999999;G[u].push_back((edge){u, v, h, w});G[v].push_back((edge){v, u, h, w});}int lim;scanf("%d %d %d", &s, &e, &lim);int l = 0, r = lim, ans = -1, len;while(l <= r){int mid = (l + r) >> 1;Dijkstra(mid);if(dis[e] != 999999999){ans = mid;len = dis[e];l = mid + 1;}else{r = mid - 1;}}if(cas++)puts("");printf("Case %d:\n", cas);if(ans == -1){puts("cannot reach destination");continue;}printf("maximum height = %d\n", ans);printf("length of shortest route = %d\n", len);}return 0;}

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