Poj3259--Wormholes（Spfa 判负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36836		Accepted: 13495

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.

Source

USACO 2006 December Gold

RE: 农场分为几块， 有的地方存在虫洞, (有负权值（边权值为时间）的边)。 问能不能回到过去。  Spfa 判 负环。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int INF = 0x3f3f3f3f; 
int dis[550], vis[550], used[550];   //used[]数组判负环； 
int n, m, t;
struct Edge
{
    int from, to, w, next;
} edge[5050];
int head[550], cnt;
void Add(int a, int b, int w)
{
    Edge E = {a, b, w, head[a]};
    edge[cnt] = E;
    head[a] = cnt++;
}
bool Spfa(int src)
{
     queue<int> q;
    for(int i = 1; i <= n; i++)
        dis[i] = INF; 
    dis[src] = 0; vis[src] = 1;
    q.push(src);    
    while(!q.empty())
    {
    //    printf("1\n");
        int u = q.front();
        q.pop(); 
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dis[v] > dis[u] + edge[i].w)
            {
                dis[v] = dis[u] + edge[i].w;
                if(!vis[v])
                {
                    vis[v] = 1;
                    q.push(v);
                    used[v]++;
                }
            }
            if(used[v] > n)     //存在负环；
                return true;
        }
    }
    return false;
}
int main()
{
    int nt;
    scanf("%d", &nt);
    while(nt--)
    {
        scanf("%d %d %d", &n, &m, &t);
        int u, v, w; cnt = 0;
        memset(vis, 0, sizeof(vis));
        memset(used, 0 , sizeof(used));
        memset(head, -1, sizeof(head));
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            Add(u, v, w);
            Add(v, u, w);
        }
        for(int i = 1; i <= t; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            Add(u, v, -w);
        } 
        if(Spfa(1))
            printf("YES\n");
        else
            printf("NO\n"); 
    }
    return 0;
}