I - Wormholes

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题目：

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 backT 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

题意：

昆虫洞

给你一个F,代表有F组测试数据。

第1行，3个数，N,M,W。

第2行到M+1行，每行3个数S,E,T，表示一条从S到E的双向路，通过这条路需要时间T,两块地之间至多有一条路。

第M+2到M+W+1行，每行3个数S,E,T，表示一条从S到E的单向路，通过这条路需要时间-T,即时间倒退T;

求从某块地出发，通过一些路和昆虫洞后还回原点，时间比他出发的时间早，使他能碰到出发前的他自己。

思路：

从题目就可以知道，这是一道最短路问题，而且还有负权边，所以应该用Bellman—Ford算法，同时还要判断是否存在负权边，判断方法在代码中讲。

代码如下：

#include<stdio.h>
#include<string.h>
#define inf 999999999int t,n,m,k,s;
int a,b,c,flag;
int v[6001],u[6001],w[6001];
int dis[6001];void init()//初始化；
{for(int i=2; i<=n; i++)dis[i]=inf;return ;
}void panduan()//判断是否存在负权边；
{for(int i=1; i<=s; i++){if(dis[v[i]]>dis[u[i]]+w[i])flag=1;}return ;
}void Bellman_ford()
{int i,j;init();dis[1]=0;flag=0;for(i=1; i<=n; i++){for(j=1; j<=s; j++){if(dis[v[j]]>dis[u[j]]+w[j])dis[v[j]]=dis[u[j]]+w[j];}}panduan();//如果进行了N-1次的放缩后还能够放缩，那么就存在负权边；if(flag==1)printf("YES\n");elseprintf("NO\n");return ;
}int main()
{scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&k);int i,j;s=1;for(i=1; i<=m;i++)//注意是双向路；{scanf("%d%d%d",&a,&b,&c);u[s]=a;v[s]=b;w[s++]=c;v[s]=a;u[s]=b;w[s++]=c;}for(i=1; i<=k; i++)//注意是单向路；{scanf("%d%d%d",&a,&b,&c);u[s]=a;v[s]=b;w[s++]=-c;}Bellman_ford();}return 0;
}

 

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