poj-3259-Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 45555 Accepted: 16812

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

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.

Source

USACO 2006 December Gold

题意：就是问这个人有没有可能在通过时空之门后看到之前的自己，即求负回路

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f;


struct  node
{
int u, v, val;
} e[5210];


int n, m, w, cnt;
int dis[510];


void Add(int u, int v, int val)
{
e[cnt].u = u;
e[cnt].v = v;
e[cnt++].val = val;
}


int main()
{
int F;
scanf("%d", &F);
while (F--)
{
cnt = 0;
scanf("%d %d %d", &n, &m, &w);
int u, v, val;
while (m--)
{
scanf("%d %d %d", &u, &v, &val);
Add(u, v, val);
Add(v, u, val);
}
while (w--)
{
scanf("%d %d %d", &u, &v, &val);
Add(u, v, -val);
}
int i,j;
int flag = 0;
memset(dis, 0, sizeof(dis));
for( i = 0;i < n; i++)
        {
            for( j = 0; j < cnt; j++)
            {
                if(dis[e[j].v] > dis[e[j].u] + e[j].val)
                {
                    dis[e[j].v] = dis[e[j].u] + e[j].val;
                     if(i == n - 1)
                            flag = 1;
                }
            }
        }


if(flag)
            printf("YES\n");
        else
            printf("NO\n");
}
return 0;
}