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 = ;
 const int MAXM = ;
 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[ * 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, ), mem(dis, inf), mem(num, );
     Q.push(st);
     vis[st] = ;
     num[st] = ;
     dis[st] = ;
     while(!Q.empty())
     {
         int a = Q.front();
         Q.pop();
         vis[a] = ;
         for(int i = head[a]; i != -; 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] = ;
                     Q.push(to);
                     num[to] ++;
                     if(num[to] > n)
                     {
                         flag = ;
                         break;
                     }
                 }
             }
         }
         if(flag)
             break;
     }
     if(flag)
         printf("YES\n");
     else
         printf("NO\n");
 }

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

POJ3259