POJ1988 Cube Stacking(并查集的应用)

最新推荐文章于 2022-03-03 13:43:07 发布

Feynman1999

最新推荐文章于 2022-03-03 13:43:07 发布

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分类专栏： Union Find 文章标签： poj

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本文介绍了一种使用并查集解决游戏模拟问题的方法。通过维护每个集合的元素个数及元素间的相对位置，实现了对立方体堆叠游戏中移动与计数操作的有效处理。

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problem

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

思路

考虑它的操作类似于集合的合并，所以考虑并查集。
在基础并查集的基础上加入两个数组：一个记录该集合中元素的个数；一个记录该集合中在i盒子上面的盒子个数。询问时相减（再减一）即可得到答案。

代码示例

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int maxn=30010;

int father[maxn];
int r[maxn];//当前编号盒子上方的盒子数(到根的距离)
int mx[maxn];//集合元素数

int p;
void init()
{
    for(int i=1;i<maxn;++i){
        father[i]=i;;
        mx[i]=1;
        r[i]=0;
    }
}

int getfather(int x)
{
    //if(x!=father[x]) return father[x]=getfather(father[x]);
    int fx=father[x];
    if(father[x]!=x){
        fx=getfather(father[x]);
        r[x]+=r[father[x]];//在压缩路径时，累加到根的距离
    }
    return father[x]=fx;
}

void Union(int x,int y)
{
    int xx=getfather(x);
    int yy=getfather(y);
    if(xx!=yy){
        father[yy]=xx;
    }
    r[yy]+=mx[xx];//到根的距离增加
    mx[xx]+=mx[yy];//数量增加
}

int main()
{
    ios::sync_with_stdio(false);
    char ch;
    int u,v;
    while(cin>>p)
    {
        init();
        while(p--)
        {
            cin>>ch;
            if(ch=='C'){
                int f;
                cin>>u;
                f=getfather(u);
                cout<<mx[f]-r[u]-1<<endl;
            }
            else{
                cin>>u>>v;
                Union(u,v);
            }
        }
    }
    return 0;
}