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 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
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;
}