poj3259 Bellman_Ford算法

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34465		Accepted: 12585

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.

!@$!#%^#$^@#%$^&*^!@$#@#$%^&**(&^%!^&*&!

#include<stdio.h>
#include<string.h>

const int INF=99999999;

struct Node{
int u;
int v;
int o;
};

Node  edge[10250];
int dis[550];
int n,m,w,nn;

void relax(int u,int v,int o)
{
if(dis[v]>dis[u]+o)
dis[v]=dis[u]+o;
}

bool Bellan_Ford()
{
int i,j,k;
for(i=1;i<=n;i++)
dis[i]=INF;

for(i=1;i<=n-1;i++)
{
for(j=1;j<=nn;j++)
relax(edge[j].u,edge[j].v,edge[j].o);
}

for(i=1;i<=nn;i++)
{
if(dis[edge[i].v]>dis[edge[i].u]+edge[i].o)
{
return false;
}
}
return true;
}

int main()
{
int F,i,j,k,flg;
int s,e,t;
scanf("%d",&F);
while(F--)
{
flg=0;nn=1;
scanf("%d %d %d",&n,&m,&w);
for(i=1;i<=m;i++)
{
scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);
nn++;
edge[nn].u=edge[nn-1].v,edge[nn].v=edge[nn-1].u,edge[nn].o=edge[nn-1].o;
nn++;
}
for(i=1;i<=w;i++)
{
scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);
edge[nn].o=-edge[nn].o;
nn++;
}
nn--;
if(Bellan_Ford()==false)
{
flg=1;
}
if(flg==1)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}

View Code