Poj 3259（bellman_ford）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25394 Accepted: 9083

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 toF (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 toE 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
用bellman_ford判负环即可，代码留存。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<cmath>#include<queue>#include<iostream>using namespace std;const int maxn = 100000 + 5;const int INF = 1000000000;typedef long long LL;typedef pair<int,int> P;int n,m,w;struct Edge{    int from,to,dis;}e[maxn];int d[maxn];int cnt;bool Bellman_ford(){    memset(d,0,sizeof(d));    for(int i = 0;i < n;i++){        for(int j = 0;j < cnt;j++){            Edge ee = e[j];            if(d[ee.to] > d[ee.from] + ee.dis){                d[ee.to] = d[ee.from] + ee.dis;                if(i == n-1) return true;            }        }    }    return false;}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d%d%d",&n,&m,&w);        cnt = 0;        for(int i = 0;i < m;i++){            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            e[cnt++] = Edge{x,y,z};            e[cnt++] = Edge{y,x,z};        }        for(int i = 0;i < w;i++){            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            e[cnt++] = Edge{x,y,-z};        }        if(Bellman_ford()) printf("YES\n");        else printf("NO\n");    }    return 0;}