题干：

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 Tseconds 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：

有若干个虫洞，给出了若干普通路径和其所用时间以及虫洞的路径和其倒回的时间，现问你能否回到出发之前的时间，注意普通路径是双向的，虫洞是单向的。

解题报告：

由题目所给信息已经可以构建一张完整的图了，然后进一步理解题目的意思其实是这张图是否存在负环，因此使用Bellman_Ford或者spfa即可。

AC代码：（邻接表储存图）（266ms）

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,cnt;
const int MAX = 505;
const int INF = 0x3f3f3f3f;
int dis[MAX],maze[MAX][MAX],cntt[MAX],head[MAX];
bool vis[MAX];
struct Edge {int to,w,ne;Edge(){}//没有此构造函数不能写  node t  这样Edge(int to,int w,int ne):to(to),w(w),ne(ne){}//可以写node(pos,cost)这样} e[200000 + 5];//数组别开小了 
void add(int u,int v,int w) {e[cnt].to = v;e[cnt].w = w;e[cnt].ne = head[u];head[u] = cnt++;
}
bool spfa(int s){dis[s]=0; vis[s]=1;  //队列初始化,s为起点int v;queue<int > q;q.push(s);while (!q.empty()){   //队列非空v=q.front();  //取队首元素q.pop();vis[v]=0;   //释放队首结点，因为这节点可能下次用来松弛其它节点，重新入队for(int i=head[v]; i!=-1; i=e[i].ne)  //对所有顶点{if (dis[e[i].to]>dis[v]+e[i].w) {  dis[e[i].to] = dis[v]+e[i].w;   //修改最短路if (vis[e[i].to]==0) {  //如果扩展结点i不在队列中，入队cntt[e[i].to]++;vis[e[i].to]=1;q.push(e[i].to);if(cntt[e[i].to] >=n) return true;}}}}return false;
}void init() {cnt = 0;memset(vis,0,sizeof(vis));memset(maze,0,sizeof(maze));memset(e,0,sizeof(e));memset(dis,INF,sizeof(dis));memset(cntt,0,sizeof(cntt));memset(head,-1,sizeof(head)); 
}
int main()
{int t;int M,W,u,v,w;cin>>t;	while(t--) {init();scanf("%d%d%d",&n,&M,&W);for(int i = 1; i<=M; i++) {scanf("%d%d%d",&u,&v,&w);add(u,v,w);add(v,u,w);}for(int i = 1; i<=W; i++) {scanf("%d%d%d",&u,&v,&w);add(u,v,-w);}cntt[1] =1;if(spfa(1)) printf("YES\n");else printf("NO\n");}return 0 ;
}

AC代码2：（邻接矩阵）（需要判重边！）（1079ms）

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
const int MAX = 505;
const int INF = 0x3f3f3f3f;
int dis[MAX],maze[MAX][MAX],cnt[MAX];
bool vis[MAX];
bool spfa(int s){
//	for(int i=0; i<=n; i++) dis[i]=99999999; //初始化每点i到s的距离dis[s]=0; vis[s]=1;  //队列初始化,s为起点int i, v;queue<int > q;q.push(s);while (!q.empty()){   //队列非空v=q.front();  //取队首元素q.pop();vis[v]=0;   //释放队首结点，因为这节点可能下次用来松弛其它节点，重新入队for(i=1; i<=n; i++)  //对所有顶点if (maze[v][i]!=INF && dis[i]>dis[v]+maze[v][i]){  dis[i] = dis[v]+maze[v][i];   //修改最短路if (vis[i]==0){  //如果扩展结点i不在队列中，入队cnt[i]++;vis[i]=1;q.push(i);if(cnt[i] >=n) return true;}}}return false;
}void init() {memset(vis,0,sizeof(vis));memset(maze,INF,sizeof(maze));memset(dis,INF,sizeof(dis));memset(cnt,0,sizeof(cnt));
}
int main()
{int t;int M,W,u,v,w;cin>>t;	while(t--) {init();scanf("%d%d%d",&n,&M,&W);for(int i = 1; i<=M; i++) {scanf("%d%d%d",&u,&v,&w);if(w<maze[u][v])maze[u][v] = maze[v][u] = w;}for(int i = 1; i<=W; i++) {scanf("%d%d%d",&u,&v,&w);maze[u][v] = -w;}cnt[1] =1;if(spfa(1)) printf("YES\n");else printf("NO\n");}return 0 ;
}

 AC代码3：Bellman_Ford算法（125ms）

//Bellman_Ford算法试试
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 3000*2
#define INF 0xFFFFFFFint t , n , m, w;
int dis[MAXN];
struct Edge{int x;int y;int value;
}e[MAXN];bool judge(){for(int i = 0 ; i < m*2+w ; i++){if(dis[e[i].y] > dis[e[i].x] + e[i].value)return false;}return true;
}void Bellman_Ford(){dis[1] = 0;for(int i = 2 ; i <= n ; i++)dis[i] = INF;for(int i = 1 ; i <= n ; i++){for(int j = 0 ; j < m*2+w; j++){if(dis[e[j].y] > dis[e[j].x] + e[j].value)dis[e[j].y] = dis[e[j].x] + e[j].value;}}if(judge())printf("NO\n");elseprintf("YES\n");
}int main()
{int uu,vv,ww,i;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&w);for(i=0;i<m*2;){scanf("%d%d%d",&uu,&vv,&ww);e[i].x=uu;e[i].y=vv;e[i++].value=ww;e[i].x=vv;e[i].y=uu;e[i++].value=ww;}for(;i<m*2+w;i++){scanf("%d%d%d",&uu,&vv,&ww);e[i].x=uu;e[i].y=vv;e[i].value=-ww;}Bellman_Ford();}return 0;
}

总结：

    1.脑残了吧调了一下午+一晚上bug尝试了邻接矩阵邻接表，发现是写了memset(dis,INF,sizeof(INF));这一行乐色？呵呵

    2.如果用邻接表储存这题别忘了去重一下不然会wa的，至于为什么权值为负的时候不需要判重，大概是因为这题只要是负的，都不影响结果？因为这题要看的是，是否存在负环啊。