POJ-3259 Wormholes（判断负环、模板）

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


题目大意：N条双向边，W条单向虫洞，走过每条边花费的时间是正的，走虫洞会让时间倒流，每条边和虫洞的花费已知，问小明从起点开始转一圈再回到起点的时间，能不能是在出发之前。
题目解析：用bellman_ford算法，判断是不是存在负环。

代码如下：

# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
using namespace std;
const int INF=1<<29;
struct edge
{
    int fr,to,w,nxt;
};
edge e[8000];
int head[550],n,cnt,dis[550];
void add(int u,int v,int w)
{
    e[cnt].fr=u;
    e[cnt].to=v;
    e[cnt].w=w;
    //e[cnt].nxt=head[u];
    //head[u]=cnt++;
    ++cnt;
}
bool bellman_ford()
{
    int i,j;
    fill(dis,dis+n+1,INF);
    dis[1]=0;
    for(i=0;i<n;++i){
        for(j=0;j<cnt;++j){
            if(dis[e[j].fr]!=INF&&dis[e[j].to]>dis[e[j].fr]+e[j].w){
                dis[e[j].to]=dis[e[j].fr]+e[j].w;
                if(i==n-1)
                    return true;
            }
        }
    }
    return false;
}
int main()
{
    int T,s,t;
    scanf("%d",&T);
    int a,b,c;
    while(T--)
    {
        cnt=0;
        memset(head,-1,sizeof(head));
        scanf("%d%d%d",&n,&s,&t);
        while(s--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        while(t--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        if(bellman_ford())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}