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：
 1 #include<stdio.h>
 2 #include<queue>
 3 #include<string.h>
 4 #define mem(a, b) memset(a, b, sizeof(a))
 5 const int MAXN = 550;
 6 const int MAXM = 3000;
 7 const int inf = 0x3f3f3f3f;
 8 using namespace std;
 9 
10 int n, m, k; //n个点 m条双向边 k个虫洞（单向边） 
11 int head[MAXN], cnt;
12 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 
13 int dis[MAXN], flag;
14 queue<int> Q;
15 
16 struct Edge
17 {
18     int to, next, w;
19 }edge[2 * MAXM];
20 
21 void add(int a, int b, int c)
22 {
23     cnt ++;
24     edge[cnt].to = b;
25     edge[cnt].w = c;
26     edge[cnt].next = head[a];
27     head[a] = cnt;
28 }
29 
30 void spfa(int st)
31 {
32     while(!Q.empty())    Q.pop();
33     mem(vis, 0), mem(dis, inf), mem(num, 0);
34     Q.push(st);
35     vis[st] = 1;
36     num[st] = 1;
37     dis[st] = 0;
38     while(!Q.empty())
39     {
40         int a = Q.front();
41         Q.pop();
42         vis[a] = 0;
43         for(int i = head[a]; i != -1; i = edge[i].next)
44         {
45             int to = edge[i].to;
46             if(dis[to] > dis[a] + edge[i].w)
47             {
48                 dis[to] = dis[a] + edge[i].w;
49                 if(!vis[to])
50                 {
51                     vis[to] = 1;
52                     Q.push(to);
53                     num[to] ++;
54                     if(num[to] > n)
55                     {
56                         flag = 1;
57                         break;
58                     }
59                 }
60             }
61         }
62         if(flag)
63             break;
64     }
65     if(flag)
66         printf("YES\n");
67     else
68         printf("NO\n");
69 }
70 
71 int main()
72 {
73     int T;
74     scanf("%d", &T);
75     while(T --)
76     {
77         cnt = 0, flag = 0, mem(head, -1);
78         scanf("%d%d%d", &n, &m, &k);
79         for(int i = 1; i <= m; i ++)
80         {
81             int a, b, c;
82             scanf("%d%d%d", &a, &b, &c);
83             add(a, b, c);
84             add(b, a, c);
85         }
86         for(int i = 1; i <= k; i ++)
87         {
88             int a, b, c;
89             scanf("%d%d%d", &a, &b, &c);
90             add(a, b, -c);
91         }
92         spfa(1);
93     }
94     return 0;
95 }
POJ3259

 

 
posted @ 2019-09-07 10:59  缘未到  阅读(143)  评论(0)  编辑  收藏  举报