HDU【2812】Building Block

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Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4269    Accepted Submission(s): 1334


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
6M 1 6C 1M 2 4M 2 6C 3C 4
 

Sample Output
102
 
#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int maxn = 30005;int fa[maxn];//父亲结点int low[maxn];//该点以下的砖块int high[maxn];//包含该砖块的并查集中包括的砖块总和void init(){    for(int i = 0 ; i < maxn; i++)    {        fa[i] = i;        low[i] = 0;        high[i] = 1;    }}int Find(int x){    if(x != fa[x])    {        int root = Find(fa[x]);         //递归的时候更新low数组,需要细细体会。        low[x] += low[fa[x]];        return fa[x] = root;    }    else        return  x;}int main(){    int P,x,y;    char c;    scanf("%d%*c",&P);    init();    while(P--)    {        scanf("%c",&c);        if(c == 'M')        {            scanf("%d%d%*c",&x,&y);                    //后面的%*c是为了消除换行符            int fx = Find(x),fy = Find(y);            if(fx == fy) continue;            fa[fx] = fy;            low[fx] = high[fy];            high[fy] += high[fx];        }        else if(c == 'C')        {            scanf("%d%*c",&x);            Find(x);            cout << low[x] << endl;        }    }    return 0;}


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