POJ1988_Cube Stacking

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 18973 Accepted: 6605Case 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

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

Sample Output

102

Source

USACO 2004 U S Open

 

题目大意：

有N(N<=30,000)堆方块，开始每堆都是一个方块。方块编号1 – N。 有两种操作：

• M x y ： 表示把方块x所在的堆，拿起来叠放到y所在的堆上。
• C x : 问方块x下面有多少个方块。

操作最多有 P (P<=100,000)次。对每次C操作，输出结果。

 

思路：

使用并查集，每加入两块砖头，让下边的砖头作为父节点，上边的砖头作为根节点；两堆砖头合并的时候，让下边堆的根节点作为上边堆的根节点。这其中，除了father数组外，还需要两个数组sum(用来记录每堆共有多少砖块)，under数组(用来记录第i块砖头下边有多少个砖头)，其中sum数组只在合并堆的时候更新，under数组在合并堆的时候和路径压缩的时候都要更新。

 

代码：

# include<iostream># include<stdio.h># include<string.h>using namespace std;const int MAXN = 30010;int father[MAXN],sum[MAXN],under[MAXN];int find(int a){    if(a==father[a])        return a;    int t = find(father[a]);//找到a的根节点    under[a] += under[father[a]];//找根节点的路径上  更新a堆下边砖块数目    father[a] = t;//更新根节点    return father[a];}void Merge(int a,int b)//将a所在的堆叠放到b所在的堆上边{    int pa = find(a);    int pb = find(b);    if(pa==pb)        return;    father[pa] = pb;//将上边的堆的根结点更新为下边的堆的根节点    under[pa] = sum[pb];    //上边的堆增加b堆的数量，这里 = 和 += 效果一样，因为上边的堆最下边也就是a的根under值为0    sum[pb] += sum[pa];    //b所在堆的数目加上a堆数目}int main(){    int P, a, b;    char cmd[20];    for(int i = 0; i <= MAXN; i++)    {        father[i] = i;        under[i] = 0;        sum[i] = 1;    }    while(~scanf("%d", &P))    {        scanf("%s",cmd);        if(cmd[0]=='M')        {            scanf("%d%d", &a, &b);            Merge(a,b);        }        else        {            scanf("%d", &a);            find(a);            printf("%d\n",under[a]);        }    }    return 0;}