POJ 1988 Cube Stacking【并查集的简单应用 堆木块】

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 15951		Accepted: 5448
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

Source

USACO 2004 U S Open

题意：给你一定数目的堆（PS：堆的数目不确定，但是不超过最大值 30,000）

            初始时每个堆都有且仅有一个方块 ，其编号安装所在堆的编号编为 1 ... 30,000 。

            输入一个数 P，表示有 P 个操作：

            C     x              计算编号为 x 的方块下面有多少个方块，输出结果。

            M     x         y   将编号为 x 的方块所在的堆放到编号为 y 所在的堆的上面。

算法：并查集。

思路【盗版的，惭愧】：

          以堆底为根【重要】（PS：因为方便用堆底的下标记录当前堆的总方块数）

          p[] 记录父节点

          r[] 记录当前方块下面的方块个数

          sum[] 记录当前堆的方块总个数，记录堆底【根】的即可。

         并查集模板：

         1.找根节点：注意向下更新时 r[] 要加上父亲的 r[]

         2.合并：因为是前者放在后者上面，又是应该以堆底为根，所以 p[fx] = fy;

                       相应要更新 r[fx] = num[fy]; // x 的根【堆底】压 y 所在堆的所有方块块

                                           num[fy] += num[fx]; // 被压的堆的方块总数更新

  

G

Accepted

484 KB

250 ms

C++

1173 B

2013-04-10 16:48:07

#include<cstdio>

const int maxn = 30000+10;

int p[maxn];
int r[maxn]; //下面的个数
int sum[maxn]; //总个数

int find(int x)
{
    if(x == p[x]) return x;

    int t = p[x];
    p[x] = find(p[x]);
    r[x] = r[x]+r[t]; //自己下面的+父亲下面的

    return p[x];
}

void Union(int x, int y)
{
    int fx = find(x);
    int fy = find(y);

    if(fx == fy) return;

    p[fx] = fy; //以堆底为根
    r[fx] = sum[fy]; //上面的堆底压了 y所在堆的所有木块
    sum[fy] = sum[fy]+sum[fx]; //更新当前堆的木块个数 保存在根中
}

void set()
{
    for(int x = 1; x < maxn; x++)
    {
        p[x] = x;
        r[x] = 0;
        sum[x] = 1;
    }
}
int main()
{
    int p;
    while(scanf("%d%*c", &p) != EOF)
    {
        set();
        char c;
        int x,y;
        while(p--)
        {
            scanf("%c", &c);
            if(c == 'M')
            {
                scanf("%d%d%*c", &x, &y);
                Union(x, y);
            }
            else if(c == 'C')
            {
                scanf("%d%*c", &x);
                find(x); //可能没有更新 r[] 输出前，更新根节点 
                printf("%d\n", r[x]);
            }
        }
    }
    return 0;
}