poj 3259 Wormholes (Bellman-ford)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39574		Accepted: 14541

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

考察bellman-ford的用法，由于写时不注意导致出现细节错误，找了半天，所以写代码时 一定要注意细节问题

题意：有n块田，m条路，w个虫洞，其中m条路是双向的，虫洞是单向的，通过虫洞可以是时间倒流，问能否回到出发以前的时间

即判断是否出现负权图；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int from;
int to;
int vis;
}t[11000];
int dis[1100],n,m,w,all;
bool bellman()
{
bool flag ;
for(int i=0;i<n-1;i++)
{
flag=false;
for(int j=0;j<all;j++)
if(dis[t[j].to]>dis[t[j].from]+t[j].vis)
{
dis[t[j].to]=dis[t[j].from]+t[j].vis;
flag=true;
}
if(!flag)
break;
}
for(int k=0;k<all;k++)
if(dis[t[k].to]>dis[t[k].from]+t[k].vis)
return true;
return false;
}
int main()
{
int F,a,b,c,i;
scanf("%d",&F);
while(F--)
{
memset(dis,0,sizeof(dis));
scanf("%d%d%d",&n,&m,&w);

all=0;
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
t[all].from=a;
t[all].to=b;
t[all++].vis=c;
t[all].from=b;
t[all].to=a;
t[all++].vis=c;
}
for(i=0;i<w;i++)
{
scanf("%d%d%d",&a,&b,&c);
t[all].from=a;
t[all].to=b;
t[all++].vis=-c;//权值为负
}
if(bellman())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}