poj3259: Wormholes(BF模板题）

http://poj.org/problem?id=3259

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.

 

题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前

解题思路：其实给出了坐标，这个时候就可以构成一张图，然后将回到从前理解为是否会出现负权环，用bellman-ford就可以解出了

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define N 900001
struct node
{
    int u,v,w;
}q[];
int dis[];
int n,m,w1,count=;
int B()
{
    int flag=;
    for(int i=;i<=n;i++)
        dis[i]=N;
    dis[]=;
    for(int i=;i<=n-;i++)
    {
        flag=;
        for(int j=;j<count;j++)
        {
            if(dis[q[j].v]>dis[q[j].u]+q[j].w)//这里u,v是不能颠倒的，因为
            {
               dis[q[j].v]=dis[q[j].u]+q[j].w;
               flag=;
            }
        }
       /* for(int j=0;j<n;j++)
            printf(".%d",dis[j]);*/
        if(!flag) break;
    }
    for(int i=;i<count;i++)
    {
         if(dis[q[i].v]>dis[q[i].u]+q[i].w)
            return ;
    }
    return ;
}
int main()
{
    int x,y,x1,T;
    scanf("%d",&T);
    while(T--)
    {
        count=;
        scanf("%d%d%d",&n,&m,&w1);
        while(m--)
        {
            scanf("%d%d%d",&x,&y,&x1);
            q[count].u=x;
            q[count].v=y;
            q[count++].w=x1;
            q[count].u=y;
            q[count].v=x;
            q[count++].w=x1;
        }
        while(w1--)
        {
            scanf("%d%d%d",&x,&y,&x1);
            q[count].u=x;
            q[count].v=y;
            q[count++].w=-x1;//他是有方向的
        }
        int t=B();
        if(t==) printf("YES\n");
        else printf("NO\n");
    }
    return ;
}

第二次写的：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF 0x7fffffff
using namespace std;
int n,m,k,tt;
struct node
{
    int x,y,z;
} q[];
int dis[];
void add(int xx,int yy,int zz)
{
    q[tt].x=xx;
    q[tt].y=yy;
    q[tt++].z=zz;
}
void BF()
{
    int flag;
    dis[]=;
    for(int i=; i<=n; i++)
    {
        flag=;
        for(int i=; i<tt; i++)
        {
            if(dis[q[i].y]>dis[q[i].x]+q[i].z)
            {
                dis[q[i].y]=dis[q[i].x]+q[i].z;
                flag=;
            }
        }
        if(flag==) break;
    }
    if(flag==) printf("YES\n");
    else printf("NO\n");
}
int main()
{
    int T,zz,xx,yy;
    cin>>T;
    while(T--)
    {
        cin>>n>>m>>k;
        tt=;
        for(int i=; i<m; i++)
        {
            cin>>xx>>yy>>zz;
            add(xx,yy,zz);
            add(yy,xx,zz);
        }
        for(int i=; i<k; i++)
        {
            cin>>xx>>yy>>zz;
            add(xx,yy,-zz);
        }
        BF();
    }
    return ;
}