poj Wormholes 【最短路径】【bellman_ford】

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.

 

【题意】有f组数据，n个场所，m条路，w个虫洞，经过虫洞连接的路后可以时光倒流。求是否可以从任意点出发后回到原点

【思路】flody算法只能用于权值是正数的最短路径，这里的虫洞的权值为负，

 

AC代码：

#include<cstdio> 
#define MAXM 2710
#define MAXV 505
#define inf 1<<29

struct node{
    int x,y,t;
}edge[MAXM];

int n,m,w;//n区域个数 m条路 w个虫洞 

int bellman_ford()
{
    int i,j,d[MAXV],flag=1,cnt=1;
    for(i=1;i<=n;i++) 
        d[i]=inf;//出发点到i的距离全部初始化为无穷大 
    while(flag)
    {
        flag=0;//是为了把所有路走一遍 
        if(cnt++>n) 
            return 1;
        //i是路的标号 
        for(i=1;i<=m;i++)//普通道路（非虫洞）找最短路 
        {
            //分别找到edge[i].x和edge[i].y的最短距离 
            if(d[edge[i].x]+edge[i].t < d[edge[i].y])  
                {d[edge[i].y] = d[edge[i].x]+edge[i].t;     flag=1;}
            if(d[edge[i].y]+edge[i].t < d[edge[i].x]) 
                {d[edge[i].x] = d[edge[i].y]+edge[i].t;        flag=1;}
        }
        for(;i<=m+w;i++)//虫洞找最短路 
            if(d[edge[i].y] > d[edge[i].x]-edge[i].t) 
                {d[edge[i].y] = d[edge[i].x]-edge[i].t;        flag=1;}
    }
    return 0;
}

int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(i=1;i<=m+w;i++)
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].t);
        if(bellman_ford()) 
            printf("YES
");
        else 
            printf("NO
");
    }
    return 0;
}