POJ 3259 Wormholes 贝尔曼福特算法判负环

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32449		Accepted: 11791

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 to F (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 to E 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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

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.

Source

USACO 2006 December Gold

题目大意：John有几个农场，农场之间有路，也有虫洞，虫洞可以让时光倒流，John从某一点出发，看看是否能回到原来的位置并且能看到出发时的自己。意思就是让时光倒流，能回到他出发前。。。。判负环。。。

又是一个判负环的题。。。。。

#include <stdio.h>
#include <string.h>

struct node
{
    int u;
    int v;
    int w;
} ls[10010];

int num;
const int inf = 99999999;

void creat(int s,int e,int t)
{
    ls[num].u = s;
    ls[num].v = e;
    ls[num].w = t;
    num++;
}

bool bf(int n)
{
    bool flag;
    int dis[710];
    for(int i = 0; i < n; i++)
        dis[i] = inf;
    for(int i = 1; i <= n - 1; i++)
    {
        flag = false;
        for(int j = 0; j < num; j++)
        {
            int u = ls[j].u;
            int v = ls[j].v;
            int w = ls[j].w;
            if(dis[v] > dis[u] + w)
            {
                flag = true;
                dis[v] = dis[u] + w;
            }
        }
        if(!flag)
            break;
    }
    flag = false;
    for(int i = 0; i < num; i++)
    {
        int u = ls[i].u;
        int v = ls[i].v;
        int w = ls[i].w;
        if(dis[v] > dis[u] + w)
        {
            flag = true;
            break;
        }
    }
    return flag;
}

int main()
{
    int n,m,k,s,e,t,w;
    while(~scanf("%d",&k))
    {
        while(k--)
        {
            num = 0;
            scanf("%d%d%d",&n,&m,&w);
            for(int i = 0; i < m; i++)
            {
                scanf("%d%d%d",&s,&e,&t);
                creat(s,e,t);
                creat(e,s,t);
            }
            for(int i = 0; i < w; i++)
            {
                scanf("%d%d%d",&s,&e,&t);
                creat(s,e,-t);
            }
            if(bf(n))
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}