poj 3259 Wormholes[ bellman_ford 判负环]

Wormholes

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.

思路：普通路双向，正值；虫洞单向，负值；bellman_ford 判负环；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef long long LL;
const int N=1002;

int dist[N];
int gra[N][N],vis[N][N];

bool bellman_ford(int n,int s)
{
    for(int i=1;i<=n;i++)
    {
        dist[i]=INF;
    }
    dist[s]=0;
    for(int i=1;i<=n-1;i++)
    {
        int flag=false;
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                if(vis[j][k]==1&&dist[k]>dist[j]+gra[j][k])
                {
                    dist[k]=dist[j]+gra[j][k];
                    flag=true;
                }
            }
        }
        if(!flag) return false;
    }

    for(int j=1;j<=n;j++){
        for(int k=1;k<=n;k++){
            if(vis[j][k]==1&&dist[k]>dist[j]+gra[j][k])
                return true;
        }
    }
    return false;
}

int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        int n,m,w;
        scanf("%d%d%d",&n,&m,&w);
        mem(vis,0);
        int s,e,t;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            if(vis[s][e])
            {
                if(gra[s][e]>t)
                    gra[s][e]=gra[e][s]=t;
            }
            else{
                gra[s][e]=gra[e][s]=t;
                vis[s][e]=vis[e][s]=1;
            }
        }
        for(int i=0;i<w;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            gra[s][e]=0-t;
            vis[s][e]=1;

        }

        int flag=0;
        for(int i=1;i<=1;i++)       //只用判1，就行了，图是联通的，不然临接矩阵超时！！
        {
            if(bellman_ford(n,i))
            {
                flag=1;
                break;
            }
        }
//        for(int i=1;i<=n;i++)
//            printf("%d___",dist[i]);
//        printf("%d\n",gra[3][1]);
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}