Poj 3259 (bellman_ford)

Wormholes

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:25394		Accepted:9083

Description

While processing his own 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
ComprisesN(1 ≤N≤ 500) fields conveniently numbered 1 ..N,
M(1 ≤M≤ 2500) paths, andW(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
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. FFarm 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
SAnd EThat requires TSeconds 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 STo
EThat also moves the traveler back TSeconds.

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 cocould travel back in time by the cycle 1-> 2-> 3-> 1, arriving back at his starting location 1 second before he leaves. he cocould start from anywhere on the cycle to accomplish this.

Source

USACO 2006 December Gold
Use bellman_ford to determine the negative ring and keep the code.

#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;}