HDU 1988 Cube Stacking （数据结构-并检查集合）

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Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 18834		Accepted: 6535
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

1
0
2

Source

USACO 2004 U S Open

题目大意：

有N个立方体和N个格子，1~N编号，一開始i立方体在i号格子上，每一个格子刚好1个立方体。如今m组操作，M a b表示将a号立方体所在的格子的所有立方体放在b号立方体所在的格子的所有立方体上面。C x表示询问x号立方体以下的立方体的个数。

解题思路：

在并查集的基础上。仅仅须要知道x到父亲的距离以及父亲究竟的距离就知道x究竟的距离。

解题代码：

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn=31000;
int father[maxn],cnt[maxn],dis[maxn];

int find(int x){
    if(father[x]!=x){
        int tmp=father[x];
        father[x]=find(father[x]);
        dis[x]+=dis[tmp];
    }
    return father[x];
}

void combine(int x,int y){
    father[x]=y;
    dis[x]+=cnt[y];
    cnt[y]+=cnt[x];
}

int main(){
    int m;
    scanf("%d",&m);
    for(int i=0;i<maxn;i++){
        father[i]=i;
        cnt[i]=1;
        dis[i]=0;
    }
    while(m-- >0){
        char ch;
        cin>>ch;
        if(ch=='M'){
            int a,b;
            scanf("%d%d",&a,&b);
            if(find(a)!=find(b)) combine(find(a),find(b));
        }else{
            int x;
            scanf("%d",&x);
            find(x);
            printf("%d\n",dis[x]);
        }
    }
    return 0;
}

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