POJ 3259 Bellman_Ford

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32474 Accepted: 11808

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

/*Source CodeProblem: 3259User: Grant YuanMemory: 224KTime: 125MSLanguage: C++Result: Accepted*/    Source Code    #include<iostream>    #include<algorithm>    #include<cstdio>    #include<cstring>    using namespace std;    int n,m,w;    struct edges    {        int from,to,cost;    };    int d[507];    edges edge[6000];    int all=0;    bool Bellman_Ford()    {        bool flag;        memset(d,0,sizeof(d));        for(int i=1;i<=n;i++)        {            flag=false;            for(int k=0;k<all;k++)            {                if(d[edge[k].to]>d[edge[k].from]+edge[k].cost){                    flag=true;                    d[edge[k].to]=d[edge[k].from]+edge[k].cost;                }            }            if(i==n&&flag) return true;        }        return false;    }    int main()    {      //  freopen("in.txt","r",stdin);        int f;int a,b,c;        scanf("%d",&f);        while(f--){            memset(edge,0,sizeof(edge));            all=0;            scanf("%d%d%d",&n,&m,&w);            for(int i=0;i<m;i++)            {                scanf("%d%d%d",&a,&b,&c);                edge[all].from=a;edge[all].to=b;edge[all++].cost=c;                 edge[all].from=b;edge[all].to=a;edge[all++].cost=c;            }              for(int i=0;i<w;i++)            {                scanf("%d%d%d",&a,&b,&c);                edge[all].from=a;edge[all].to=b;edge[all++].cost=-c;            }            bool  ans=Bellman_Ford();            if(ans) printf("YES\n");            else printf("NO\n");        }        return 0;    }