（最短路 spfa）Wormholes -- poj -- 3259

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

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37356		Accepted: 13734

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

 

 

代码：

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
const int INF = (1<<30)-1;
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define N 5500

int n, m, w, dist[N], G[N][N], vis[N];
int u;

struct node
{
    int v, t, next;
} a[N];


int Head[N], cnt;

void Init()
{
    cnt = 0;
    memset(Head, -1, sizeof(Head));
    memset(vis, 0, sizeof(vis));
    for(int i=0; i<=n; i++)
    {
        dist[i] = INF;
        for(int j=0; j<=i; j++)
            G[i][j] = G[j][i] = INF;
    }
}

void Add(int u, int v, int t)
{
    a[cnt].v = v;
    a[cnt].t = t;
    a[cnt].next = Head[u];
    Head[u] = cnt++;
}

int spfa()
{
    queue<int>Q;
    Q.push(1);
    dist[1] = 0;
    vis[1] = 1;

    while(Q.size())
    {
        int u = Q.front();  Q.pop();

        vis[u] = 0;

        for(int i=Head[u]; i!=-1; i=a[i].next)
        {
            int v = a[i].v;
            int t = a[i].t;
            if(dist[u] + t < dist[v])
            {
                dist[v] = dist[u] + t;
                if(vis[v] == 0)
                {
                    Q.push(v);
                    vis[v] = 1;
                }
            }
        }

        if(dist[1] < 0) ///当 dist[1] 为负数的时候说明它又回到了原点
            return 1;
    }
    return 0;
}


int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int i, u, v, x;

        scanf("%d%d%d", &n, &m, &w);

        Init();
        for(i=1; i<=m; i++)
        {
            scanf("%d%d%d", &u, &v, &x);
            Add(u, v, x);
            Add(v, u, x);
        }

        for(i=1; i<=w; i++)
        {
            scanf("%d%d%d", &u, &v, &x);
            Add(u, v, -x);
        }

        int ans = spfa();

        if(ans)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}