HDU 2818 Building Block 带权并查集

Building Block

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5966 Accepted Submission(s): 1853

Problem 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

Source

2009 Multi-University Training Contest 1 - Host by TJU

题意：

有n块砖和p次操作：

M x y:表示将编号为x所在的这一列砖移到编号为y在的那一列砖的上面，如果两块砖已经在同一列了，就跳过。

C x：表示询问编号为x的砖的下面有多少砖。

坑点：砖的编号实际为0~30000 - 1

分析：

还是关于一些集合合并的问题，只不过这次要询问下面有多少块砖。所以只是单纯的并查集是不够的。

我们用rk[x]表示编号为x的砖下面有多少块砖。将在上面的砖连到它下面的砖上。则路径压缩时合并：rk[x]+=rk[fa[x]];

对于需要合并的砖a、b：

不难发现，对于移动到上方的砖a，现在a所在的那一列的每一块砖的rk都需要加上siz[b]，

//siz[i]表示i所在的这一列有多少块砖

可以直接在root a上加上siz[b]，注意合并后siz的改变。

至于查询，可以再查找一下它现在的根，如果之前有根的改变，那么在查找的过程中就会将改变的值加上。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}
return x*f;
}

const int N = 30000 + 10;
int p;
int fa
,rk
,siz
;

int find(int x){
if(x==fa[x]) return fa[x];
int d=find(fa[x]);
rk[x]+=rk[fa[x]];
return fa[x]=d;
}

void link(int x,int y){
int fx=find(x),fy=find(y);
if(fx==fy) return ;
rk[fx]+=siz[fy];
siz[fy]+=siz[fx];
fa[fx]=fy;
}

int main(){
p=read();
for(register int i=0;i<=N-10;i++) siz[i]=1,fa[i]=i,rk[i]=0;
for(register int j=1;j<=p;j++){
char s[5];scanf("%s",s);
if(s[0]=='M'){
int x=read(),y=read();
link(x,y);
}
else{
int x=read();find(x);
printf("%d\n",rk[x]);
}
}
return 0;
}