并查集+路径压缩(poj1988)

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

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 19122		Accepted: 6664
Case Time Limit: 1000MS

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

Sample Output

题意：可以把题目中的意思抽象为栈集合的合并，题目中有两个操作

M 把包含x的栈放在包含y的栈的上面；

C统计包含x的栈中x下面有多少个元素；

分析：此题是并查集路径压缩的典型题目，与前面的种类并查集还有些不同；题目中用到f[],num[],dis[]三个数组，f数组记录i的父节点，num数组记录以i为根的集合有多少个元素，dis记录i到根节点的距离，那么i下面的元素就是x=finde(i)；num[x]-dis[i]-1个元素;

程序：

#include"stdio.h"

#include"string.h"

#include"iostream"

#include"map"

#include"string"

#include"queue"

#include"stdlib.h"

#include"math.h"

#define M 30009

#define eps 1e-10

#define inf 1000000000

#define mod 2333333

using namespace std;

int f[M],num[M],dis[M];

int finde(int x)

{

    if(x!=f[x])

    {

        int t=f[x];

        f[x]=finde(f[x]);

        dis[x]+=dis[t];

    }

    return f[x];

}

void make(int a,int b)

{

    int x=finde(a);

    int y=finde(b);

    if(x!=y)

    {

        f[y]=x;

        dis[y]+=num[x];

        num[x]+=num[y];

    }

}

int main()

{

    int n,i,a,b;

    char str[3];

    while(scanf("%d",&n)!=-1)

    {

        for(i=1;i<=M+1;i++)

        {

            f[i]=i;

            num[i]=1;

            dis[i]=0;

        }

        while(n--)

        {

            scanf("%s",str);

            if(str[0]=='M')

            {

                scanf("%d%d",&a,&b);

                make(a,b);

            }

            else

            {

                scanf("%d",&a);

                int x=finde(a);

                printf("%d\n",num[x]-dis[a]-1);

            }

        }

    }

}

巴特西

并查集+路径压缩(poj1988)

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