poj-3259-Wormholes-BF

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在本文中，我们跟随农民FJ的脚步，探索他的农场中的奇妙虫洞。通过详细的地图描述，我们了解了如何在一个农场内，从一个地点出发，经过路径和虫洞的组合，回到出发点之前的时间。文章深入探讨了使用Bellman-Ford算法来解决这个问题的策略，并提供了代码实现。读者将了解到如何通过调整路径权重和应用动态规划技巧，实现时间旅行的可能。

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

Sample Output

NO
YES

两种路径，一种双向的，一种单向的，把单向的变成负数，然后用Bellman-Ford，判断负全环，另外，数组开大点

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX 0x3f3f3f
int n, m,k;
int cont = 0;
struct node
{
    int u, v, w;
} ls[5100];
int d[1500];
void BF()
{
    int i, j;
    for(i = 1; i <= n; i++)
        d[i] = MAX;
    d[1] = 0;
    bool flag ;
    for(i = 0; i < n-1; i++)
    {
        flag = false;
        for(j = 0; j < cont; j++)
        {
            int x = ls[j].u, y = ls[j].v;
            if(d[y] > d[x]+ls[j].w)
            {
                d[y] = d[x]+ls[j].w;
                flag = true;
            }

        }
        if(!flag)
            break;
    }
    flag = false;
    for(j = 0; j < cont; j++)
    {
        int x = ls[j].u, y = ls[j].v;
        if(d[y] > d[x]+ls[j].w)
        {
            printf("YES\n");
            flag = true;
            break;
        }
    }
    if(!flag)
        printf("NO\n");
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        cont = 0;
        for(int i = 0; i < m; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            ls[cont].u = u;
            ls[cont].v = v;
            ls[cont++].w = w;
            ls[cont].u = v;
            ls[cont].v = u;
            ls[cont++].w = w;
        }
        for(int i = 0; i < k; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            w = -w;
            ls[cont].u = u;
            ls[cont].v = v;
            ls[cont++].w = w;

        }
        BF();
    }
    return 0;
}