• 带权值得并查集


    Building Block

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 151   Accepted Submission(s) : 57

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    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:

      M x y把x的管放在y管的上面

     C x求x的下面有多少个砖块

    down[i]表示i到根节点之间有多少块砖,再维护一个sum[x]数组,表示管x上一共串有多少个砖块

    find每次更新down数组;

    #include<iostream>
    #include<stdio.h>
    using namespace std;
    int par[30005];
    int sum[30005];//上面有多少;
    int down[30005];
    int findi(int x)
    {
        if(x==par[x])
            return x;
        int p=findi(par[x]);
        down[x]=down[par[x]]+down[x];
        return par[x]=p;
    }
    void unioni(int x,int y)
    {
        int xx=findi(x);
        int yy=findi(y);
        if(xx!=yy)
        {
            par[xx]=yy;
            down[xx]=sum[yy];
            sum[yy]=sum[yy]+sum[xx];
    
        }
    
    
    }
    
    int main()
    {
    int n;
    for(int i=0;i<30005;i++)
        sum[i]=1;
    for(int i=0;i<30005;i++)
        down[i]=0;
    for(int i=0;i<30005;i++)
         par[i]=i;
    scanf("%d",&n);
    
    while(n--)
    {
        char a;
        cin>>a;
        if(a=='M')
        {
            int b,c;
           scanf("%d%d",&b,&c);
            unioni(b,c);
        }
        if(a=='C')
        {
            int d;
             scanf("%d",&d);
             findi(d);
             printf("%d
    ",down[d]);
        }
    }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/2014slx/p/7220836.html
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