题目地址

http://poj.org/problem?id=3259

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

这题主要是把题读懂。题意是：

2个test case 每个test case 第一行：  
N M W  
N个点 M条双向正权边 W条单向负权边    第一个test case 最后一行 3 1 3 是单向负权边，3->1的边权值是-3 

因为有负边，所以用bellman-ford算法就行。

AC代码：

#include <stdio.h>
#include <string.h>
#define N 505
#define inf 100000
int d[N];
int n, m;
typedef struct
{
    int from;
    int to;
    int cost;
}node;
node edges[5300];
int bellman_ford(int m)
{
    int i, j, k, t;
    int ok;
    for(i=1; i<=n; i++)
        d[i] = inf;
    d[1] = 0;
    for(i=1; i<=n-1; i++)
    {
        ok = 1;
        for(j=1; j<=m; j++)
        {
            if(d[edges[j].from] > d[edges[j].to] + edges[j].cost)
            {
                d[edges[j].from] = d[edges[j].to] + edges[j].cost;
                ok = 0;
            }
        }
        if(ok)
            break;
    }
    for(j=1; j<=m; j++)
    {
        if(d[edges[j].from] > d[edges[j].to] + edges[j].cost)
            return 0;
    }
    return 1;
}

int main()
{
    int i, j, t, x;
    int a, b, c, w;


    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d%d", &n, &m, &w);
        memset(edges, 0, sizeof(edges));
        for(i=0, j=1; j<=m;j++)
        {
            scanf("%d%d%d", &a, &b, &c);
            i++;
            edges[i].from = a;
            edges[i].to= b;
            edges[i].cost = c;
            i++;
            edges[i].from = b;
            edges[i].to = a;
            edges[i].cost = c;

        }
        for(j=1; j<=w; j++)
        {
            scanf("%d%d%d", &a, &b, &c);

            i++;
            edges[i].from = a ;
            edges[i].to = b;
            edges[i].cost = -c;

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