poj--3259

---恢复内容开始---

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：John是一个农场主，他的农场出现了虫洞并可以进行时空穿梭，给出农场的虫洞地图，问John是否可以看到原来的自己

题解：本题可以看作一个单源最短路径问题，首先，可以从一个点开始出发，到达余下的所有节点，其次，这个题目还有负数的权值边，然后就可以用Bellman-Ford算法来求能否

　　　　看到自己了。

Bellman-Ford算法：

　　从单个源点出发，到达余下的节点。这个算法本质是一个动态规划的算法，在算法导论一书中，讲到一个松弛操作就是先对路径的最大距离进行初始化

　　for(i=1;i<=n;i++)　{dist[i]=INF;}　　dist[1]=0;

　　算法的实现：

　　

bool bellman_ford()
{
    for(i=1;i<=n;i++)
        dist[i]=INF;dist[1]=0;
    for(i=1;i<=n;i++)
    {
        bool update=true;
        for(j=1;j<m;j++)
        {
            if(dist[edge[i].to]>dist[edge[i].from+edge[i].cost)
            {
                dist[edge[i].to]=dist[edge[i].from]+edge[i].cost;//更新dist距离
                   update=false;
            }
        }
        if(update) return false;    
    }
    for(i=0;i<m;i++)
     {
       if(dist[edge[i].to]>dist[edge[i].from+edge[i].cost)//判断是否存在负数环
        return true;
    }    
  return false;  
}    

 

代码：

　　

#include<iostream>
using namespace std;
int F,N,M,W;
int S,E,T;
struct EDGE{
    int from,to;
    int cost;
};
const int INF=100000;
EDGE edge[10000];
int dist[510];
int m;
bool bellman_ford()
{
    for(int i=1;i<=N;i++)
    {
        dist[i]=INF;
    }
    dist[1]=0;
    for(int i=1;i<N;i++)
    {
        bool update=true;
        for(int j=0;j<m;j++)
        {
            int u=edge[j].from;
            int v=edge[j].to;
            int t=edge[j].cost;
            
            if(dist[v]>dist[u]+t)
            {
                dist[v]=dist[u]+t;
                update=false;
            }
        }
        if(update)    return false;
    }
    for(int i=0;i<m;i++)
    {
        if(dist[edge[i].to]>dist[edge[i].from]+edge[i].cost)
            return true;
    }
    return false;
}
int main()
{
    while(cin>>F)
    {
        while(F--)
        {
            scanf("%d%d%d",&N,&M,&W);
            int u,v,t;
            m=0;
            for(int i=1;i<=M;i++)
            {
                scanf("%d%d%d",&u,&v,&t);
                edge[m].from=u;
                edge[m].to=v;
                edge[m++].cost=t;
                
                edge[m].from=v;
                edge[m].to=u;
                edge[m++].cost=t;
            }
            for(int i=1;i<=W;i++)
            {
                scanf("%d%d%d",&u,&v,&t);
                edge[m].from=u;
                edge[m].to=v;
                edge[m++].cost=-t;
            }
            if(bellman_ford())    printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}