POJ3259 Wormholes 【spfa判负环】

题目链接：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:75598		Accepted: 28136

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.

题目大意：n 个点, m 条双向边，代表经过所需的时间，w 条单向边，代表经过所减少的时间。问该图中是否存在负环。

思路：

1.简单的判环，用spfa即可，在不存在环的情况下，任意点在进行路径松弛时，最多被其他的点更新一次。那么任意点的最多入队次数只能是 n 次。当存在任何一点的入队次数大于顶点数n，即说明存在环。

2.判负环，即路径往小松弛。判正环时，即路径往大更新。注意dis数组初始化即可。

代码如下：

#include<stdio.h>
#include<queue>
#include<string.h>
#define mem(a, b) memset(a, b, sizeof(a))
const int MAXN = 550;
const int MAXM = 3000;
const int inf = 0x3f3f3f3f;
using namespace std;

int n, m, k; //n个点 m条双向边 k个虫洞（单向边） 
int head[MAXN], cnt;
int vis[MAXN], num[MAXN];//num表示第i个点的入队次数 用来判断负环是否存在 
int dis[MAXN], flag;
queue<int> Q;

struct Edge
{
    int to, next, w;
}edge[2 * MAXM];

void add(int a, int b, int c)
{
    cnt ++;
    edge[cnt].to = b;
    edge[cnt].w = c;
    edge[cnt].next = head[a];
    head[a] = cnt;
}

void spfa(int st)
{
    while(!Q.empty())    Q.pop();
    mem(vis, 0), mem(dis, inf), mem(num, 0);
    Q.push(st);
    vis[st] = 1;
    num[st] = 1;
    dis[st] = 0;
    while(!Q.empty())
    {
        int a = Q.front();
        Q.pop();
        vis[a] = 0;
        for(int i = head[a]; i != -1; i = edge[i].next)
        {
            int to = edge[i].to;
            if(dis[to] > dis[a] + edge[i].w)
            {
                dis[to] = dis[a] + edge[i].w;
                if(!vis[to])
                {
                    vis[to] = 1;
                    Q.push(to);
                    num[to] ++;
                    if(num[to] > n)
                    {
                        flag = 1;
                        break;
                    }
                }
            }
        }
        if(flag)
            break;
    }
    if(flag)
        printf("YES
");
    else
        printf("NO
");
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        cnt = 0, flag = 0, mem(head, -1);
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= m; i ++)
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            add(a, b, c);
            add(b, a, c);
        }
        for(int i = 1; i <= k; i ++)
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            add(a, b, -c);
        }
        spfa(1);
    }
    return 0;
}