POJ3259 Wormholes 【spfa判负环】
作者：互联网
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
标签：cnt,num,POJ3259,int,vis,判负,spfa,edge,dis 来源： https://www.cnblogs.com/yuanweidao/p/11479589.html