POJ 3259 - Wormholes(Bellman负环)

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本文探讨了如何利用Bellman算法检测是否存在时光倒流的可能性，即判断图中是否存在负权回路，通过解决一个有趣的农场时空旅行问题来阐述算法原理。

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Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 46424 Accepted: 17159
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.

题意：
有n个点，m个双向路径，w个虫洞（虫洞可以退回一定时间），问能否通过一定方式实现时光倒流（判断是否有负环）。

解题思路：
用bellman判断负环。如果途中不存在从s可达的负圈，那么最短路就不会经过同一个点两次（也就是说，最多到达V-1条边）。

AC代码：

#include<stdio.h>
#include<memory.h>
const int maxn = 5205;
int dis[maxn];
struct node{int from,to,cost;}edge[maxn];
int n,m,w;
bool bellman()
{
    memset(dis,0,sizeof(dis));
    for(int i = 1;i <= n;i++)
    {
        for(int j = 1;j <= (m<<1)+w;j++)
        {
            node e = edge[j];
            if(dis[e.to]>dis[e.from]+e.cost)
            {
                dis[e.to] = dis[e.from]+e.cost;
                if(i == n)  return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(int i = 1;i <= m<<1;i+=2)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            edge[i].from=a,edge[i].to=b,edge[i].cost=c;
            edge[i+1].from=b,edge[i+1].to=a,edge[i+1].cost=c;
        }
        for(int i = (m<<1)+1;i <= (m<<1)+w;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            edge[i].from=a,edge[i].to=b,edge[i].cost=(-c);
        }
        printf("%s\n",bellman()?"YES":"NO");
    }
    return 0;
}