POJ 3259 Wormholes（Bellman Ford 单源最短路径（可求带负权边的））

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
//第一个Bellman Ford 算法。感觉不熟悉、对这个算法的实现不是怎么理解
//这题是看了别人的代码才会写的 ,Path双边，虫洞单向，现在还不明白为什么随便选择个边作起点都可以AC
//这题是利用 Bellman Ford 判断有从起点S可到达的负回路
#include

 <string.h> #include <stdio.h> 
#include <iostream> 
#define N 6666 
#define Max 0xfffffff 
using namespace std; 
struct node
 { 
int s,e,t; 
}Edge[N]; int d[503]; int n,m,w; 
bool bellman_ford() { 
int i,j,temp; d[1]=0; 
for(i=2;i<=n;i++) 
d[i]=Max; for(i=1;i<n;i++) 
for(j=1;j<=w;j++) {
 temp=Edge[j].t+d[Edge[j].s]; 
if(d[Edge[j].s]!=Max&&d[Edge[j].e]>temp)
 d[Edge[j].e]=temp; 
} 
for(i=1;i<=w;i++) 
if(d[Edge[i].s]!=Max&&d[Edge[i].e]>Edge[i].t+d[Edge[i].s]) 
return 0; 
return 1; 
} 
int main() { 
// freopen("in.txt","r",stdin); 
int i,j,t; scanf("%d",&t); 
while(t--) { 
j=0; 
scanf("%d%d%d",&n,&m,&w); 
for(i=1;i<=m;i++) { 
++j; 
scanf("%d%d%d",&Edge[j].s,&Edge[j].e,&Edge[j].t); 
++j; 
Edge[j].s=Edge[j-1].e; 
Edge[j].e=Edge[j-1].s; 
Edge[j].t=Edge[j-1].t; 
} for(i=1;i<=w;i++) { 
++j; scanf("%d%d%d",&Edge[j].s,&Edge[j].e,&Edge[j].t);
 Edge[j].t=-Edge[j].t; 
} w=j; 
if(!bellman_ford()) 
printf("YES\n"); 
else printf("NO\n"); 
} 
return 0; 
}