POJ3259——Bellman_foed——Wormholes

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

大意：有F个test，共N个编号，M个正向的u,v,t，即从u到v要花t秒，可双向，W个虫洞反向的，只单向，要花-t秒，问最后能不能到达出发之前。

就有if(d[edge.v]>d[edge.u]+edge.t){

       flag = 0;

      d[edge.v] = d[edge.u]+edge.t;

}

意为如果虫洞返回后的时间前于出发的时间，那么满足，经过N此循环都是满足的话，那么就是满足的。

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int MAX = 22222222;

int F,N,M,W,cnt;

int d[6000];

struct edge

{

    int u;

    int v;

    int t;

}edge[6000];



bool bellman_ford()

{

    for(int i = 1; i <= N;i++)

        d[i] = MAX;

    d[1] = 0;

    bool flag;

    for(int i = 1; i <= N;i++){

            flag = 1;

       for(int j = 0; j < cnt; j++){

            if(d[edge[j].v]>d[edge[j].u]+edge[j].t){

                    flag = 0;

              d[edge[j].v] = d[edge[j].u]+edge[j].t;

         }

      }

    if(flag)

        return true;

   }

   return false;

}







int main()

{

    int u,v,t;

    scanf("%d",&F);

    while(F--){

            cnt = 0;

            scanf("%d%d%d",&N,&M,&W);

            for(int i = 1; i <= M;i++){

                scanf("%d%d%d",&u,&v,&t);

                edge[cnt].u = u; edge[cnt].v = v; edge[cnt].t = t;

                cnt++;

                edge[cnt].u = v; edge[cnt].v = u; edge[cnt].t = t;

                cnt++;

            }

            for(int i = 1; i <= W;i++){

                    scanf("%d%d%d",&u,&v,&t);

                    edge[cnt].u = u;edge[cnt].v = v;edge[cnt].t = -t;

                    cnt++;

            }

        if(bellman_ford())

            printf("NO\n");

        else printf("YES\n");

    }

    return 0;

}

View Code