ZOJ 1298 Domino Effect

首先dijkstra求出最短路,求的时候顺便求出1能到的最长路记为ans * 2.

枚举每条边,满足dis[i] + dis[j] + g[i][j] > ans 更新ans.

最后输出答案ans / 2;

/* * ZOJ 1298.cpp * *  Created on: May 27, 2013 *      Author: root */#include <iostream>#include <cstdio>#include <queue>#include <memory.h>#include <limits.h>using namespace std;const int maxn = 505;struct edge{int v, next;double w;}es[maxn * 5];struct node{int i;double v;node(int ii = 0, double vv = 0):i(ii), v(vv){}bool operator<(const node & rhs)const{return v > rhs.v;}};int head[maxn], n, m, fro, tar, type;double ans;double dis[maxn];bool vis[maxn];void addEdge(int u, int v, double w, int eidx){es[eidx].v = v;es[eidx].w = w;es[eidx].next = head[u];head[u] = eidx;}void dijkstra(){type = 0;for(int i = 1; i <= n; ++i){dis[i] = INT_MAX;}memset(vis, 0, sizeof(vis));dis[1] = 0;priority_queue<node> q;q.push(node(1, 0));while(q.size()){node t = q.top();q.pop();int u = t.i;if(vis[u])continue;vis[u] = 1;if(t.v >= ans){fro = u;ans = t.v;}for(int ne = head[u]; ne != -1; ne = es[ne].next){int v = es[ne].v;double w = es[ne].w;if(dis[v] > dis[u] + w && !vis[v]){dis[v] = dis[u] + w;q.push(node(v, dis[v]));}}}}int main(){int cas = 1;while(scanf("%d %d", &n, &m)){if(!n && !m)break;memset(head, -1, sizeof(head));int eidx = 0;for(int i = 0; i < m; ++i){int u, v;double w;scanf("%d %d %lf", &u, &v, &w);addEdge(u, v, w, eidx++);addEdge(v, u, w, eidx++);}ans = 0;fro = 1;dijkstra();ans *= 2;for(int i = 1; i <= n; ++i){for(int ne = head[i]; ne != -1; ne = es[ne].next){int u = i, v = es[ne].v;double w = es[ne].w;if(dis[u] + dis[v] + w > ans){fro = u, tar = v;type = 1;ans = dis[u] + dis[v] + w;}}}printf("System #%d\n", cas++);if(type == 0){if(m == 0) fro = 1;printf("The last domino falls after %.1lf seconds, at key domino %d.\n", ans / 2, fro);}else{printf("The last domino falls after %.1lf seconds, between key dominoes %d and %d.\n", ans / 2, fro, tar);}printf("\n");}return 0;}