poj 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 34529 Accepted: 12610

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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

给一个图，判断有没有负环。

bellman_ford一下，看能不能超n

#include <iostream>#include <stdio.h>#include <algorithm>#include <math.h>#include <string.h>#include <queue>#include <stack>#include <map>#include <set>#include <stdlib.h>#include <vector>using namespace std;#define INF 1<<30struct line{int b,e;int t;}e[6666];int minn[1000];int T,n,m,w,s,en,t,cnt;bool relax(int p){int sum=minn[e[p].b]+e[p].t;if(sum<minn[e[p].e]){minn[e[p].e]=sum;return true;}return false;}bool bellman(){for(int i=1;i<=n;i++) minn[i]=INF;bool flag;for(int i=1;;i++){if(i>n) return true;flag=false;for(int j=1;j<=cnt;j++){if(relax(j)) flag=true;}if(!flag) return false;}}int main(){cin>>T;while(T--){cnt=0;cin>>n>>m>>w;        for(int i=1;i<=m;i++){cin>>s>>en>>t;cnt++;e[cnt].b=s;e[cnt].e=en;e[cnt].t=t;cnt++;e[cnt].b=en;e[cnt].e=s;e[cnt].t=t;}for(int i=1;i<=w;i++){cin>>s>>en>>t;cnt++;e[cnt].b=s;e[cnt].e=en;e[cnt].t=-t;}if(bellman()) cout<<"YES"<<endl;else cout<<"NO"<<endl;}return 0;}