hdu 2818 Building Block

Building Block

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

Total Submission(s): 3986 Accepted Submission(s): 1235



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

[align=left]Output[/align]
Output the count for each C operations in one line.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1
0
2

[align=left]Source[/align]
2009 Multi-University Training Contest 1 - Host by TJU

带权并查集

#include<stdio.h>
#include<string.h>
#define M 30000+10
int x[M],rank[M],up[M];//rank记录该点下边有几个点，up记录该点上边有几个
void init(){
for(int i=0;i<M;i++){
x[i]=i; rank[i]=0; up[i]=1; //up初始化为1，以便后边操作
}
}
int find(int k){
int temp=x[k];
if(x[k]==k) return k;
x[k]=find(x[k]);
rank[k]+=rank[temp];//子节点下边的点的个数等于本身的个数加上父节点下边的个数
return x[k];
}
void merge(int a,int b)
{
int fa=find(a);
int fb=find(b);
if(fa==fb) return;
x[fa]=fb;
rank[fa]=up[fb];//合并时，fa下边几个就等于fb上边有几个因为刚开始初始化为1所以不用再加1
up[fb]+=up[fa];//更新
}
int main()
{
int p,a,b,c,i;
char ch[5];
while(scanf("%d",&p)!=EOF){
init();
for(i=0;i<p;i++){
scanf("%s",ch);
if(ch[0]=='M'){
scanf("%d%d",&a,&b);
merge(a,b);
}else{
scanf("%d",&c);
int k=find(c);
printf("%d\n",rank[c]);
}
}
}
return 0;
}