POJ 3259 Wormholes (最短路)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34302		Accepted: 12520

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.

 

 

 

以每个点为起点跑一次，检测是否有负环。

#include <iostream>
#include <cstdio>
using    namespace    std;

const    int    INF = 0xfffffff;
const    int    SIZE = 5500;
int    D[505];
int    N,M,W,F;
struct    Node
{
    int    from,to,cost;
}G[SIZE];

bool    Bellman_Ford(int);
bool    relax(int,int,int);
int    main(void)
{
    scanf("%d",&F);
    while(F --)
    {
        scanf("%d%d%d",&N,&M,&W);
        int    i = 0;
        while(i < 2 * M)
        {
            scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
            i ++;
            G[i] = G[i - 1];
            swap(G[i].from,G[i].to);
            i ++;
        }
        while(i < W + M * 2)
        {
            scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
            G[i].cost = -G[i].cost;
            i ++;
        }
        bool    flag = true;
        for(int i = 1;i <= N;i ++)
            if(!Bellman_Ford(i))
            {
                puts("YES");
                flag = false;
                break;
            }
        if(flag)
            puts("NO");
    }

    return    0;
}

bool    Bellman_Ford(int s)
{
    fill(D,D + 505,INF);
    D[s] = 0;
    bool    update;

    for(int i = 0;i < N - 1;i ++)
    {
        update = false;
        for(int i = 0;i < M * 2 + W;i ++)
            if(relax(G[i].from,G[i].to,G[i].cost))
                update = true;
        if(!update)
            break;
    }
    for(int i = 0;i < M * 2 + W;i ++)
        if(relax(G[i].from,G[i].to,G[i].cost))
            return    false;
    return    true;
}

bool    relax(int from,int to,int cost)
{
    if(D[to] > D[from] + cost)
    {
        D[to] = D[from] + cost;
        return    true;
    }
    return    false;
}