Wormholes(SPFA+Bellman)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36860		Accepted: 13505

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
题解：
这道题意思是这个人的农田里有道路，还有虫洞，道路是双向的，虫洞单向，他喜欢虫洞旅行，想要从起点再回到起点时间在出发的时间之前，经过虫洞时间倒退；
做这道题其实就是判断最短路有没有负环的问题；因为只要有负环，就会无限循环，到起点自然就时间倒退了；
代码：SPFA

 #include<stdio.h>
 #include<queue>
 #include<string.h>
 using namespace std;
 const int INF=0x3f3f3f3f;
 const int MAXN=;
 const int MAXM=*;
 int dis[MAXN],vis[MAXN],used[MAXN],head[MAXM];
 int N,W,M,en,flot;
 queue<int>dl;
 struct Edge{
     int from,to,value,next;
 };
 Edge edg[MAXM];
 void initial(){
     memset(dis,INF,sizeof(dis));
     memset(vis,,sizeof(vis));
     memset(used,,sizeof(used));
     memset(head,-,sizeof(head));
     while(!dl.empty())dl.pop();
     en=;flot=;
 }
 void print(){
     if(flot)puts("YES");
     else puts("NO");
 }
 void add(int u,int v,int w){
     Edge E={u,v,w,head[u]};
     edg[en]=E;
     head[u]=en++;
 }
 void SPFA(int sx){
     dis[sx]=;vis[sx]=;dl.push(sx);
     used[sx]++;
     while(!dl.empty()){
         int k=dl.front();
         dl.pop();
         vis[k]=;
         if(used[k]>N){
             flot=;
             break;
                 }
         for(int i=head[k];i!=-;i=edg[i].next){
             int v=edg[i].to;
             if(dis[k]+edg[i].value<dis[v]){
                 dis[v]=dis[k]+edg[i].value;
                 if(!vis[v]){
                     vis[v]=;
                     dl.push(v);
                     used[v]++;
                     if(used[v]>N){
                         flot=;return ;
                     }
                 }
             }
         }
     }
 }
 void get(){
     int F,a,b,c;
     scanf("%d",&F);
     while(F--){
         initial();
         scanf("%d%d%d",&N,&M,&W);
         while(M--){
             scanf("%d%d%d",&a,&b,&c);
             add(a,b,c);
             add(b,a,c);
         }
         while(W--){
             scanf("%d%d%d",&a,&b,&c);
             add(a,b,-c);
         }
         SPFA();
         print();
     }
 }
 int main(){
     get();
     return ;
 }

Bellman:

 #include<stdio.h>
 #include<string.h>
 const int INF=0x3f3f3f3f;
 const int MAXN=;
 const int MAXM=;
 int dis[MAXN];
 struct Edge{
     int u,v,w;
 };
 Edge edg[MAXM];
 int N,M,W,top;
 bool Bellman(int sx){
     int u,v,w;
     memset(dis,INF,sizeof(dis));
     dis[sx]=;
     for(int i=;i<=N;i++){
         for(int j=;j<top;j++){
             u=edg[j].u;v=edg[j].v;w=edg[j].w;
             if(dis[u]+w<dis[v])dis[v]=dis[u]+w;
         }
     }
     for(int i=;i<top;i++){
         u=edg[i].u;v=edg[i].v;w=edg[i].w;
         if(dis[u]+w<dis[v])return false;
     }
     return true;
 }
 int main(){
     int F;
     int a,b,c;
     scanf("%d",&F);
     while(F--){
         top=;
         scanf("%d%d%d",&N,&M,&W);
         while(M--){
             scanf("%d%d%d",&a,&b,&c);
             edg[top].u=a;edg[top].v=b;edg[top++].w=c;
             edg[top].u=b;edg[top].v=a;edg[top++].w=c;
         }
         while(W--){
             scanf("%d%d%d",&a,&b,&c);
             edg[top].u=a;edg[top].v=b;edg[top++].w=-c;
         }
         if(Bellman())puts("NO");
         else puts("YES");
     }
     return ;
 }