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
Hint
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.
/*
判断负权回路
加开一个数组
use[]
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
struct node
{
int x,y,n;
}e[10000];
int k[1000];
int d[1000];
bool v[1000];
int f[10000000];
int use[1000];
int tot;
int F;
int N,M,W;
void add(int a,int b,int c)
{
e[++tot].x=b;
e[tot].y=c;
e[tot].n=k[a];
k[a]=tot;
}
bool SPFA(int start)
{
int head=0,tail=1;
f[tail]=start;
d[start]=0;
v[start]=true;
while (head<tail)
{
int x=f[++head];
v[x]=false;
int t=k[x];
while (t)
{
if (d[e[t].x]>(d[x]+e[t].y))
{
d[e[t].x]=d[x]+e[t].y;
if (!v[e[t].x])
{
f[++tail]=e[t].x;
v[e[t].x]=true;
use[e[t].x]++;
if (use[e[t].x]>N)
{
return true;
}
}
}
t=e[t].n;
}
}
return false;
}
int main()
{
scanf("%d\n",&F);
for (int i=1;i<=F;i++)
{
tot=0;
memset(v,0,sizeof(v));
memset(e,0,sizeof(e));
memset(k,0,sizeof(k));
memset(d,63,sizeof(d));
memset(use,0,sizeof(use));
scanf("%d %d %d\n",&N,&M,&W);
for (int j=1;j<=M;j++)
{
int a,b,c;
scanf("%d %d %d\n",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
for (int j=1;j<=W;j++)
{
int a,b,c;
scanf("%d %d %d\n",&a,&b,&c);
add(a,b,-c);
}
bool ans=false;
for (int j=1;j<=N;j++)
{
if (d[j]>1000000)
{
if (SPFA(j))
{
ans=true;
break;
}
}
}
if (ans) printf("YES\n");else printf("NO\n");
}
}
