Bellman algorithm PKU 3259

Wormholes

Time limit:2000 ms		Memory limit:65536 K
Total submissions:31593		Accepted:11497

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 mapsF(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.

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include <limits.h>#include <ctype.h>#include <string.h>#include <string>#include <math.h>#include <algorithm>#include <iostream>#include <queue>#include <stack>#include <deque>#include <vector>#include <set>//#include <map>using namespace std;const int VM = 520;const int EM = 5020;const int INF = 0x3f3f3f3f;//const int INF = 0x3f3f3f3f;struct Edge{    int u,v;    int w;}edge[EM<<1];int N,M,W,cnt,dis[VM];void addage(int cu,int cv,int cw){    edge[cnt].u = cu;    edge[cnt].v = cv;    edge[cnt].w = cw;    cnt++;}int Bellman(){    int i,j,flag;    for(i=0;i<N;i++){        dis[i] = INF;    }    for(i=1;i<N;i++){        flag = 1;        for(j=0;j<cnt;j++){            if(dis[edge[j].u] > dis[edge[j].v]+edge[j].w){                dis[edge[j].u] = dis[edge[j].v]+edge[j].w;                flag = 0;            }        }        if(flag){            break;        }    }    for(i=0;i<cnt;i++){        if(dis[edge[i].u] > dis[edge[i].v]+edge[i].w){            return 1;        }    }    return 0;}int main(){    int t;    int i;    while(~scanf("%d",&t)){        while(t--){            scanf("%d%d%d",&N,&M,&W);            cnt = 0;            int u,v,w;            for(i=0;i<M;i++){                scanf("%d%d%d",&u,&v,&w);                addage(u,v,w);                addage(v,u,w);            }            for(i=0;i<W;i++){                scanf("%d%d%d",&u,&v,&w);                addage(u,v,-w);            }            if(Bellman()){                printf("YES\n");            }            else{                printf("NO\n");            }        }    }    return 0;}

Bellman algorithm PKU 3259