POJ 3259 Wormholes 图论 贝尔曼-福特算法（Bellman-Ford）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25757 Accepted: 9254

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N,M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps toF (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S,E, T) that describe, respectively: A one way path from S toE that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

这道题是穿越题，穿越。。。题目意思是有个闲的无聊的农夫John（从题目中看农夫不是一般的闲），在自己农场里建了一个图，并且搞了一些正常的路和一些虫洞（就是科幻电影里面经常时光倒流什么什么的路径），并且正常的路是双向路，虫洞是单向的，问农夫从任意一个顶点出发，是否可以当回到此顶点的时候发生时光倒流，即时间小于出发的时间，此题看来还是运用Bellman-Ford算法中负权环的定义，如果图中出现一条负权环的话，则可以说明存在时光倒流的可能，输出YES，不存在负权环的时候，输出NO，这样题目就被愉快的AC了。。。

下面是AC代码：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;struct Edge{    int u, v;        int time;int flag;}edge[5500];int dist[600],n,m,w;void relax(int u, int v, int time){    if(dist[v] > dist[u] + time)        dist[v] = dist[u] + time;}bool Bellman_Ford(){int i;for(i=1;i<=n;i++)dist[i]=100000;    for(i=1; i<=n-1; ++i)        for(int j=1; j<=2*m+w; ++j)            relax(edge[j].u, edge[j].v, edge[j].time);    bool flag = 1;        for(i=1; i<=2*m+w; ++i)        if(dist[edge[i].v] > dist[edge[i].u] + edge[i].time )        {            flag = 0;            break;        }    return flag;}int main(){int i,t;scanf("%d",&t);while(t--){memset(edge,0,sizeof(edge));scanf("%d%d%d",&n,&m,&w);for(i=1;i<=2*m;i+=2){scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].time);edge[i+1].u=edge[i].v;edge[i+1].v=edge[i].u;edge[i+1].time=edge[i].time;}for(i;i<=2*m+w;i++){scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].time);edge[i].time=-edge[i].time;}if(Bellman_Ford()==0)printf("YES\n");elseprintf("NO\n");}return 0;}