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Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 66356 Accepted: 24748
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
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.
Source
USACO 2006 December Gold
解题报告（注释题解）：
#include<iostream>
using namespace std;
const int inf=10010;
struct node{
int f,t,c;
};
int S,E,T,N,M,W,NUM;
node n[5200];
int bellman_ford()
{
int dis[505];
dis[0]=0;
for(int i=1;i<N;i++)dis[i]=inf;//开始时设每一个点到别的点的距离无限大。就是每个点的权值
for(int i=1;i<N;i++)//以这个图的点得个数为外层循环，，由于是从任一点开始，所以这层循环求的是任一点的最短距离。
{
int flag=0;
for(int j=1;j<=NUM;j++)//以有向边的个数为内层循环，对每一个边进行松弛。
{
if(dis[n[j].t]>dis[n[j].f]+n[j].c)
{
dis[n[j].t]=dis[n[j].f]+n[j].c;
flag=1;//当循环进不来时说明不能松弛了。
}
}
if(!flag)break;//不能松弛就跳出循环。
}
for(int i=1;i<=NUM;i++)
{
if(dis[n[i].t]>dis[n[i].f]+n[i].c)
return true;
}
return false;
}
int main()
{
int nn;
cin>>nn;
while(nn--)
{
NUM=1;
cin>>N>>M>>W;
while(M--)
{
cin>>S>>E>>T;
n[NUM].f=S;
n[NUM].t=E;
n[NUM++].c=T;//无向边，故需要将两个方向都记下。
n[NUM].f=E;
n[NUM].t=S;
n[NUM++].c=T;
}
while(W--)
{
cin>>S>>E>>T;
n[NUM].f=S;
n[NUM].t=E;
n[NUM++].c=(0-T);
}
if(bellman_ford())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}