poj3259——Wormholes（Eellman-Ford算法）

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

其实就是计算连通图中是否存在负权环路，虫洞就表示负权路，正常的路是双向的，虫洞就覆盖一条单向路就行

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 5005
#define mod 1000000007
using namespace std;
int n,m,w,e;
int dis[MAXN];
struct Node
{
    int beg,end,t;
};
Node edge[MAXN<<1];
void add(int b,int en,int t)
{
    edge[e].beg=b;
    edge[e].end=en;
    edge[e].t=t;
    e++;
}
bool relax(int p)
{
    int t=dis[edge[p].beg]+edge[p].t;
    if(dis[edge[p].end]>t)
    {
        dis[edge[p].end]=t;
        return true;
    }
    return false;
}
bool bellman()
{
    for(int i=1;i<=n;++i)
        dis[i]=INF;
    dis[1]=0;
    for(int i=1; i<n; ++i)
    {
        bool flag=false;
        for(int j=0; j<e; ++j)
        {
            if(relax(j))
                flag=true;
        }
        if(dis[1]<0)
            return true;
        if(!flag)
            return false;
    }
    for(int j=0; j<e; ++j)
        if(relax(j))
            return true;
    return false;
}
int main()
{
    int f;
    scanf("%d",&f);
    while(f--)
    {
        e=0;
        scanf("%d%d%d",&n,&m,&w);
        int a,b,c;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        while(w--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        if(bellman())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}