POJ 1988 Cube Stacking (带权并查集)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 int par[int(3e4 + 5)] , cnt[int(3e4 + 5)] , dis[int(3e4 + 5)];
 6 //cnt[i]表示i所在点的大小，dis[i]表示i离最上面节点的距离(个数)
 7 int Find(int n) {
 8     if(par[n] == n)
 9         return n;
10     int temp = Find(par[n]); //先算出n的父节点的dis
11     dis[n] += dis[par[n]];
12     par[n] = temp; //路径压缩
13     return temp;
14 }
15 
16 void Union(int u , int v) {
17     int fu = Find(u) , fv = Find(v);
18     if(fu == fv)
19         return ;
20     par[fv] = fu; 
21     dis[fv] = cnt[fu]; //Find函数中dis[fv]并没有回溯增加过
22     cnt[fu] += cnt[fv]; //总个数相加
23 }
24 
25 int main()
26 {
27     int n , u , v;
28     char q[3];
29     while(~scanf("%d" , &n)) {
30         memset(dis , 0 , sizeof(dis));
31         for(int i = 1 ; i <= 3e4 ; ++i) {
32             cnt[i] = 1;
33             par[i] = i;
34         }
35         for(int i = 0 ; i < n ; ++i) {
36             scanf("%s" , q);
37             if(q[0] == ‘M‘) {
38                 scanf("%d %d" , &u , &v);
39                 Union(u , v);        
40             }
41             else {
42                 scanf("%d" , &u);
43                 int x = Find(u); //回溯累加一次
44                 printf("%d\n" , cnt[x] - dis[u] - 1);
45             }
46         }
47     }
48     return 0;
49 }