POJ1988 Cube Stacking(并查集)

题目：

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 24296 Accepted: 8517Case 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

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

Sample Output

102

Source

USACO 2004 U S Open

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代码：

#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define shu 20using namespace std;int n,fa[shu],r[shu],mx[shu];//r[x]表示x到根节点的距离，mx[x]表示集合中有多少个盒子void init(){    int i;    for(i=1; i<shu; i++)    {        fa[i]=i;        r[i]=0;        mx[i]=1;    }}int find(int x)//查找并查集祖先{    int fx=fa[x];    if(fa[x]!=x)    {        fx=find(fa[x]);        r[x]+=r[fa[x]];    }    return fa[x]=fx;}void u(int x,int y)//将两个集合合并{    int fx=find(x);    int fy=find(y);    fa[fy]=fx;    r[fy]+=mx[fx];//合并时更改相应的值    mx[fx]+=mx[fy];}int main(){    char op[3];    int i,j;    while(~scanf("%d",&n))    {        init();        while(n--)        {            scanf("%s",op);            if(op[0]=='C')            {                scanf("%d",&i);                int f=find(i);                printf("%d\n",mx[f]-r[i]-1);//该集合盒子个数减去在i盒子之上的个数            }            else            {                scanf("%d%d",&i,&j);                u(i,j);            }        }    }    return 0;}

 x所在集合的元素个数 - 在它上面的个数 - 它本身 就是在它之下的个数