Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 28098 | Accepted: 9858 | |
| 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
题目链接：http://poj.org/problem?id=1988
题意：有n个方块，进行m次操作。有两种操作类型，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。
带全并查集裸模板，用两个数组表示该点到根节点一共有多少个，和表示该点的下面有多少个。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 1e6+5;
int father[maxn];
int sum[maxn];
int dis[maxn];
int find(int x)
{
if(x != father[x])
{
int f = father[x];
father[x] = find(father[x]);
sum[x] += sum[f];
}
return father[x];
}
int main()
{
int m;
cin >> m;
for(int i = 1; i <= m; i++)
{
father[i] = i;
sum[i] = 0;
dis[i] = 1;
}
while(m --)
{
char op;
int x, y;
cin >> op;
if(op == 'M')
{
cin >> x >> y;
int fx = find(x);
int fy = find(y);
if(fx != fy)
{
father[fx] = fy;
sum[fx] += dis[fy];
dis[fy] += dis[fx];
}
}
else
{
cin >> x;
int fx = find(x);
cout << sum[x] << endl;
}
}
return 0;
}