poj1988 简单并查集

B - 叠叠乐

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:30000KB 64bit IO Format:%lld & %llu

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Description

Input

Output

Sample Input

Sample Output

Hint

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

1

0

2

解析见代码：

/*

poj1988 ,并查集

题目大意：初始时有N块木块，编号1-N，分开放置，

然后对这些木块开始执行P条操作，M a,b是将a所在的

木块堆移到b所在的木块堆上,C a是查询编号为a的木块在

其之下有多少个木块

思路分析：如果查询的是a号木块所在的木块堆一共有多

少堆，那么这道题就是简单的维护下集合内元素数量就可以，但是现在

问的是它下面有多少块，所以还需要维护的就是它下面的木块数目

初始化under[i]=0，然后对于under数组的更新有两个地方，对于

一个子集的根节点（最下面的木块）在merge时就可以更新了，其他

的木块则是要在压缩路径的时候进行更新，under[i]=under[father[i]]

注意一定要先更新后 压缩路径，要不然就没有更新的效果

*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

const int maxn=+;

int father[maxn];

int sum[maxn];

int under[maxn];

int findroot(int x)

{

    if(x==father[x]) return x;

    int t=findroot(father[x]);

    under[x]+=under[father[x]];//先更新under数组,后压缩路径

    father[x]=t;

    return father[x];

}

void merge(int a,int b)//根是在下面的木块

{

    int pa=findroot(a);

    int pb=findroot(b);

    if(pa==pb) return;

    father[pb]=pa;

    under[pb]=sum[pa];

    sum[pa]+=sum[pb];

}

int main()

{

    for(int i=;i<=maxn;i++)

    {

        father[i]=i;

        sum[i]=;

        under[i]=;

    }

    int p;

    scanf("%d",&p);

    char s[];

    int a,b;

    while(p--)

    {

        scanf("%s",s);

        if(s[]=='M')

        {

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

            merge(b,a);

        }

        else

        {

           scanf("%d",&a);

           findroot(a);//祖先节点可能发生了 变化 ，因此需要进行更新

           printf("%d\n",under[a]);

        }

    }

}

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poj1988 简单并查集

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