hdoj 1272 小希的迷宫

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 #define N 100010
 5 using namespace std;
 6 int per[N], h[N];
 7 void init()
 8 {
 9     for(int i = 1; i <= N; i++)
10         per[i] = i;
11 }
12 int find(int x)
13 {
14     return x==per[x] ? x : find(per[x]);
15 } 
16 int join(int x, int y)//判断是否成环 
17 {
18     int fx = find(x);
19     int fy = find(y);
20     if(fx == fy)//因为给的两个点本来就是连通的，如果根节点一样，就成环了 
21         return 0;
22     else
23     {
24         per[fx] = fy;
25         return 1;
26     } 
27 }
28 int main()
29 {
30     int n, m;
31     while(~scanf("%d%d", &n, &m), n!=-1 && m!=-1)
32     {
33         memset(h, 0, sizeof(h));
34         if(n == 0 && m == 0)
35         {
36             printf("Yes\n");
37             continue;
38         }
39         int flag = 1;
40         h[n] = 1;
41         h[m] = 1;
42         init();
43         join(n, m);
44         while(~scanf("%d%d", &n, &m), n&&m)
45         {
46             int t = join(n, m);//用t记录是否成环 
47             if(!h[n])
48                 h[n] = 1;
49             if(!h[m])
50                 h[m] = 1;
51             if(t == 0)
52                 flag = 0;
53         }
54         int root = 0;
55         for(int i = 1; i <= N; i++)//判断根节点的个数 
56         {
57             if(h[i] && per[i] == i)
58                 root++;
59         }
60         if(flag == 0 || root > 1)
61             printf("No\n");
62         else
63             printf("Yes\n");
64         
65     }
66     return 0;
67 }