1988:Cube Stacking
总时间限制:
2000ms
单个测试点时间限制:
1000ms
内存限制:
65536kB
描述
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.
输入
* 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.
输出
Print the output from each of the count operations in the same order as the input file.
样例输入
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
样例输出
1 0 2
来源
USACO 2004 U S Open
思路：
题意说是有n块砖，编号从1到n，有两种操作，操作一（‘M’）是把含有x编号的那一堆砖放到含有编号y的那一堆砖的上面，操作二(‘C’)是查询编号为x的砖的下面有多少块砖。传统并查集+维护size和deep两个数组，find函数在路径压缩的过程中需要把深度也依次加上，到代表这里的时候就是这个集合的最大深度，合并过程中要先将被合并的个数加到合并别人的集合的深度上去，然后在合并个数，最后将被合并的集合清零。
#include<iostream>
#include<algorithm>
#include<stdio.h>
const int N =1e5+5;
using namespace std;
int deep[N];
int size[N];
int p[N];
int find(int x)
{
if(p[x]!=x)
{
int t=p[x];
p[x]=find(p[x]);
deep[x]+=deep[t];
}
return p[x];
}
void Union(int x,int y)
{
int xx=find(x);
int yy=find(y);
if(xx!=yy)
{
p[xx]=yy;
deep[xx]+=size[yy];
size[yy]+=size[xx];
size[xx]=0;
}
}
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
p[i]=i;
size[i]=1;
}
for(int i=1;i<=n;i++)
{
char op[5];
scanf("%s",op);
if(op[0]=='M')
{
int a,b;
scanf("%d %d",&a,&b);
Union(a,b);
}
else if(op[0]=='C')
{
int a;
scanf("%d",&a);
find(a);
printf("%d\n",deep[a]);
}
}
return 0;
}
该题链接：
OpenJudge - 1988:Cube Stacking
