POJ - 1988 - Cube Stacking

POJ - 1988 - Cube Stacking

Time Limit: 2000MS     Memory Limit: 30000K     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

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

Sample Output

1
0
2

题意：给你很多个，每个堆有数字编号，给你两个操作，M a b代表，吧第i堆的所有放在第b堆的上面，C a 代表请输出编号为a的堆下面有多少个堆

把其他堆往上面放，放我想到了并查集，就是要多加两个数组，一个是记录每个堆下有几个堆，用under数组表示，另外一个就是记录落起来一整块总共有多少，用sum表示

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 30005
using namespace std;

int u[MAXN], under[MAXN], sum[MAXN];

int find(int x){      
    if(x == u[x])
        return x;
    int a = find(u[x]);
    under[x]+=under[u[x]];    //在压缩路径的时候，要动态更新under数组的值
    u[x] = a;
    return u[x];
}

void un(int a, int b){     //把a所在的堆的所有放在b堆之上
    a = find(a); 
    b = find(b);
    if(a == b)  
        return;
    u[a] = b;
    under[a] = sum[b];     //*a放在了b之上，当然a的下面有sum[b]个
    sum[b]+=sum[a];        //*摞上去之后，b堆总数要加a堆总数
}

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF){
        getchar();
        for(int i = 0; i < MAXN; i++){
            u[i] = i;
            under[i] = 0;
            sum[i] = 1;
        }

        int a, b;
        char c;
        for(int i = 0; i < n; i++){
            c = getchar();
            if(c == 'M'){
                scanf("%d %d", &a, &b);
                getchar();
                un(a, b);
            }
            else if(c == 'C'){
                scanf("%d", &a);
                getchar();
                find(a);                //这一句一定不能少，最后输出的时候更新下under数组
                printf("%d\n", under[a]);
            }
        }
    }
    return 0;
}