POJ 3259 Wormholes 【spfa 负权回路】

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探讨了在一个包含特殊路径和虫洞的农场中寻找能够实现时间旅行的可能性，通过SPFA算法判断是否存在导致时间倒流的负权回路。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58541		Accepted: 21881

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.

判断是否存在负权回路；

spfa模板题，只需在每次pop时记录次数，如果次数已经大于等于n；

那么一定存在负权回路；

#include<vector>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
#define mms(x, y) memset(x, y, sizeof x)
#define MAX 1000
#define INF 0x3f3f3f3f
struct edge
{
    int to, c;
};
int times[MAX];
int vis[MAX];
int dis[MAX];
int n, m, w;
queue <int> q;
vector <edge> v[MAX];
void ADD(int x, int y, int z)
{
    v[x].push_back({y, z});
}
void add(int x, int y, int z)
{
    ADD(x, y, z);
    ADD(y, x, z);
}
void init()
{
    mms(times, 0);
    mms(vis, 0);
    for(int i = 1; i <= MAX; i++)
    {
        dis[i] = INF;
        v[i].clear();
    }
    while(!q.empty())
        q.pop();
}
bool spfa(int s)
{
    dis[s] = 0;
    q.push(s);
    vis[s] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        times[u]++;
        if(times[u] >= n)
            return 1;
        int len = v[u].size();
        for(int i = 0; i < len; i++)
        {
            int to = v[u][i].to;
            int c = v[u][i].c;
            if(dis[to] > dis[u] + c)
            {
                dis[to] = dis[u] + c;
                if(!vis[to])
                {
                    q.push(to);
                    vis[to] = 1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    int M;
    scanf("%d", &M);
    while(M--)
    {
        init();
        scanf("%d%d%d", &n, &m, &w);
        for(int k = 0, x, y, z; k < m; k++)
        {
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z);
        }
        for(int k = 0, x, y, z; k < w; k++)
        {
            scanf("%d%d%d", &x, &y, &z);
            ADD(x, y, -z);
        }
        if(spfa(1))
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}