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
题目大意：
FJ童鞋家的农田里面有若干个虫洞，有一些虫洞是相连的，可以消耗一些时间爬过去(无向)。还有一些虫洞比较特殊，从一头爬到那一头之后时间会退后一些(有向)
FJ希望通过爬虫洞(?)来时时间退后，从而能看到另外一个自己。根据输入的虫洞信息来判断是否可以实现FJ的愿望。
解题思路：
说白了就是判断有向图有没有负环。Bellman-Ford算法和SPFA算法都可以用来判断是否有负环。
1.Bellman-Ford算法：
For i= to |G.V|-
For each edge(u,v)属于G.E
RELAX(u,v,w)
For each edge(u,v)属于G.E //此循环用来判断负环
If (v.d>u.d+w(u,v)
Return FALSE;
Return TRUE;
2.SPFA算法：
可以通过判断 顶点i进入队列的次数是否大于N-1来判断是否存在负环。
PS：相对于没有负环的有N个顶点的有向图来说，一个顶点最多松弛N-1次即可达到最短路。
Code(SPFA):
#include<stdio.h>
#include<string>
#include<iostream>
#include<limits.h>
#include<queue>
using namespace std;
int edge[][],times[],dis[];
bool vis[];
int N;
void init(int N)
{
for (int i=; i<=N; i++)
for (int j=; j<=N; j++)
edge[i][j]=INT_MAX;
}
bool SPFA(int begin)
{
for (int i=; i<=N; i++)
{
dis[i]=INT_MAX;
vis[i]=;
times[i]=;
}
queue<int> Q;
Q.push(begin);
dis[begin]=;
vis[begin]=;
times[begin]++;
while (!Q.empty())
{
begin=Q.front();
Q.pop();
vis[begin]=;
for (int j=; j<=N; j++)
if (j!=begin&&edge[begin][j]!=INT_MAX&&dis[j]>dis[begin]+edge[begin][j])
{
dis[j]=edge[begin][j]+dis[begin];
if (!vis[j])
{
Q.push(j);
vis[j]=;
times[j]++;
if (times[j]>=N) return ;
}
}
}
return ;
}
int main()
{
int T;
cin>>T;
while (T--)
{
int M,W;
cin>>N>>M>>W;
init(N);
for (int i=; i<=M; i++)
{
int x1,x2,x3;
scanf("%d%d%d",&x1,&x2,&x3);
if (x3<edge[x1][x2])
edge[x1][x2]=edge[x2][x1]=x3;
}
for (int i=; i<=W; i++)
{
int x1,x2,x3;
scanf("%d%d%d",&x1,&x2,&x3);
edge[x1][x2]=-x3;
}
bool ok=SPFA();
if (ok) printf("NO\n");
else printf("YES\n");
}
return ;
}