POJ 3259 Wormholes （Bellman_ford算法）

题目链接：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 45077		Accepted: 16625

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.

 

题目大意：有一个农场有n块田地，m条路连接任意两条路（无向），w个虫洞（单向）回到t秒之前，现在需要你判断她从1到n，再从n回到1是否能遇到离开前的自己

解题思路：由于题目的特殊性，只需要将穿越虫洞的时间存为负值然后用Bellman_ford算法判断是否存在负权值回路即可 如果存在说明能看到否则反之。

Bellman_ford算法  http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.html

AC代码：

#include <stdio.h>
#include <string.h>
#define inf 9999999
int dis[550];
int p[550][550];
int n,m,w,ans;
struct edge
{
    int x,y,z;
}g[3000<<1];
void add(int s,int e,int t)
{
    g[ans].x = s;
    g[ans].y = e;
    g[ans].z = t;
    ans ++;
}
bool Bellman_ford()
{
    int i,j;
    for (i = 1; i <= n; i ++)
        dis[i] = inf;
    dis[1] = 0;

    for (i = 1; i < n; i ++)
        for (j = 0; j < ans; j ++)
            if (dis[g[j].y] > dis[g[j].x]+g[j].z)
                dis[g[j].y] = dis[g[j].x]+g[j].z;

    for (j = 0; j < ans; j ++) //遍历所有的边
        if (dis[g[j].y] > dis[g[j].x]+g[j].z)
            return false;
    return true;
}
int main ()
{
    int s,e,t,i,j,f;
    scanf("%d",&f);
    while (f --)
    {
        ans = 0;
        scanf("%d%d%d",&n,&m,&w);
        for (i = 0; i < m; i ++)
        {
            scanf("%d%d%d",&s,&e,&t);
            add(s,e,t);
            add(e,s,t);
        }
        for (i = 0; i < w; i ++)
        {
            scanf("%d%d%d",&s,&e,&t);
            add(s,e,-t);
        }
        bool f = Bellman_ford();
        if (f)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}