POJ3259 Wormholes(BF)

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分类专栏： ---- POJ ----- -------- 最短路 ------

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探讨了在一个包含多个地点、路径及特殊虫洞的农场中，是否存在一条路径使得旅行者能够回到过去的时间点。通过构建图论模型并运用Bellman-Ford算法来检测是否存在负权回路，从而判断是否能实现时间倒流。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 31457		Accepted: 11448

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

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

题目大体意思虫洞问题，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

就是求是否有负权环.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>

using namespace std;
#define INF 0x3f3f3f3f

struct node
{
    int x,y,z;
}q[100010];

int n,m,k;
int t;
int num[100001];
void add(int x,int y,int z)
{
    q[t].x = x;
    q[t].y = y;
    q[t++].z = z;
    q[t].x = y;
    q[t].y = x;
    q[t++].z = z;
}

int BF()
{
    for(int i=0;i<=n+1;i++)
    {
        num[i] = INF;
    }
    num[0]=0;
    int flag = 0;
    for(int i=0;i<n-1;i++)
    {
        flag = 0;
        for(int j=0;j<t;j++)
        {
            if(num[q[j].y]>num[q[j].x] + q[j].z)
            {
                num[q[j].y] = num[q[j].x] + q[j].z;
                flag = 1;
            }
        }
        if(flag == 0)
        {
            break;
        }
    }
    for(int i=0;i<t;i++)
    {
        if(num[q[i].y]>num[q[i].x] + q[i].z)
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        t = 0;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<m;i++)
        {
            int x, y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
        }
        for(int i=0;i<k;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            q[t].x = x;
            q[t].y = y;
            q[t++].z = -z;
        }

        if(!BF())
        {
            cout << "NO" << endl;
        }
        else
        {
            cout << "YES" << endl;
        }
    }
    return 0;
}