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POJ1988-Cube Stacking(带权并查集)

2024-07-14 23:56:02   220人阅读

Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 18858 Accepted: 6547
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：n个方块（n<=30000）,p组操作（p<=100000）,操作有两种，M a b 将含有a的堆放在包含b的堆上，还有一种是 C a统计a下面有多少个方块
思路：带权并查集，一堆中最顶上的方块作为父节点，用dist[X] 统计X到父亲节点的距离，rank[fa[X]]表示团的大小，两者相减即为答案
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 30000+10;
int fa[maxn];
int rank[maxn];
int dist[maxn];
void init(){
    for(int i = 0; i < maxn; i++){
        fa[i] = i;
        rank[i] = 1;
        dist[i] = 0;
    }
}
int find(int x){
    if(x != fa[x]){
        int t = fa[x];
        fa[x] = find(fa[x]);
        dist[x] += dist[t];
    }
    return fa[x];
}


int main(){
    int n;
    while(~scanf("%d",&n)){
        init();
        char op;
        while(n--){
            cin >> op;
            int a,b;
            if(op=='M'){
                scanf("%d%d",&a,&b);
                int faA = find(a);
                int faB = find(b);
                if(faA != faB){
                    fa[faB] = faA;
                    dist[faB] = rank[faA];
                    rank[faA] += rank[faB];
                }
            }else{
                scanf("%d",&a);
                int x = find(a);
                printf("%d\n",rank[x]-dist[a]-1);
            }
        }
    }
    return 0;
}

































        

            
                
                  
                
                

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