Cube Stacking POJ1988 【并查集的应用】

http://poj.org/problem?id=1988

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:

moves and counts.

* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.

* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a ‘M‘ for a move operation or a ‘C‘ for a count operation. For move operations, the line also contains two integers: X and Y.For
count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file.

Sample Input

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

Sample Output

1
0
2

求出任何一个节点的下属的个数

/*
带权并查集，一堆中最顶上的方块作为父节点，用dis[X] 统计X到父亲节点的距离，num[fa[X]]表示团的大小，两者相减即为答案
*/
#include <stdio.h>
#include <string.h>
#define N 30000+100//结点个数
int num[N];//记录这个团队的人数
int dis[N];//记录有个点到根节点的距离
int per[N];//表示父节点

void init()
{
    for(int i=0;i<N;++i)
    {
        num[i]=1;
        dis[i]=0;
        per[i]=i;
    }
}

int find(int x)
{
    if(x==per[x]) return x;
    int t=per[x];
    per[x]=find(per[x]);
    dis[x]+=dis[t];//当查找一个数时，
    return per[x];
}

void join(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        per[fy]=fx;
        dis[fy]=num[fx];
        num[fx]+=num[fy];
    }
}
int main()
{
    int n;
    int a,b;
    char ch;
    while(~scanf("%d",&n))
    {
        init();
        while(n--)
        {
            getchar();
            scanf("%c",&ch);
            if(ch=='M')
            {
                scanf("%d%d",&a,&b);
                join(a,b);
            }
            else if(ch=='C')
            {
                scanf("%d",&a);
                int x=find(a);
                printf("%d\n",num[x]-dis[a]-1);
            }
        }
    }
    return 0;
}

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