HDU 2818 && POJ1988 帶權並查集

Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.
C X : Count the number of blocks under block X

You are request to find out the output for each C operation.

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

Output

Output the count for each C operations in one line.

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

今天打比賽前寫的一道帶權並查集

題意   搬磚頭  把包含有 a的磚頭堆裏的 所有磚頭   轉移到  包含b的磚頭堆上    問   K磚頭下有幾塊磚頭

思路 ：並查集   額外開兩個數組    一個記錄這堆磚頭堆的磚頭高度   另一個記錄k磚頭下有幾塊磚頭      每次合併的時候更新磚頭高度    和被合併的磚頭堆   最底下磚頭   下面有幾塊磚頭    用戶find  時更新

ac code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include <cstdlib>

using namespace std;
const int maxn=30005;
int root[maxn],under[maxn],height[maxn];
void init(int n)
{
    for(int i=0;i<=n;i++)
        root[i]=i,under[i]=0,height[i]=1;
}

int fin(int x)
{
    if(root[x]==x) return x;
    int rot=root[x];
    root[x]=fin(root[x]);
    under[x]+=under[rot];//遞歸更新要查找的點
    return root[x];
}

void Union(int u,int v)
{
    int ru=fin(u);
    int rv=fin(v);
    if(ru==rv) return;
    root[ru]=rv;//ru堆的最底下變成了rv
    under[ru]=height[rv];//ru磚頭下有rv塊磚頭
    height[rv]+=height[ru];//更新rv堆磚頭高度
}

int main()
{
    char ch[5];
    int u,v,n;
    while(scanf("%d",&n)!=EOF)
    {
        init(n<30000?n:30000);
        for(int i=0;i<n;i++)
        {
            scanf("%s",ch);
            if(ch[0]=='M')
            {
                scanf("%d%d",&u,&v);
                Union(u,v);
            }
            else
            {
                scanf("%d",&u);
                fin(u);//這裏必須有   因爲算法是延遲更新的    這裏如果不查找   就不會更新
                printf("%d\n",under[u]);
            }
        }
    }
    return 0;
}

現在也找不到什麼帶權並查集的題了    也就發了兩篇博客   總的來說   帶權並查集的核心在於   延遲更新

建圖的時候理清楚合併的集合的關係就好辦了