Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 17359 | Accepted: 5991 | |
| 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.
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
大意：
给定N个方块，排成一行，将它们编号1到N。再给出P个操作：
①M i j表示将i所在的那一堆移到j所在那一堆的顶上。
②C i表示一个询问，询问i下面有多少个方块。
•你需要写一个程序来完成这些操作。
毫无疑问，这么大的数据范围，暴力肯定不行。而效率几乎为O(N)的并查集跳入了我们的视线。
我们设f[i]是每个方块所在位置的最底下方块的编号（好晕啊~），h[i]是每个方块之下有多少个方块（开始赋成0）。每次读入合并操作Y和X时：
首先找到X和Y的父亲（最底下的点），同时更新：f[fy]=fx。但是很快问题就来了——我们怎么更新f[fy]的高度呢？我们不能获知左边堆里最上面一层的h[i]!
咨询了RZZ后，我又开了个数组deep[i],表示i所在的堆内的高度（仅当i是底层的方格）。初始化时是1。进行如图的操作时,h[fy]=deep[fx]，同时deep[fx]+=deep[fy]。
有人可能要问：右侧堆中底层的元素已经转移了，那么它上面的点呢？这里，要用到类似于线段树里的lazy-tag思想，即询问时再更新值。如果询问右侧某一点，我们只需在它寻找父亲、路径压缩时，再增加一个修改h的操作即可。
以下是AC代码：
#include<stdio.h>
const int maxn=30001;
int f[maxn],h[maxn],deep[maxn];
using namespace std;
int get(int u)
{
if (u==f[u]) return u;
int temp=f[u];
f[u]=get(f[u]);
h[u]+=h[temp];
return f[u];
}
void Union(int x,int y)
{
f[x]=y;
h[x]=deep[y];
deep[y]+=deep[x];
deep[x]=0;
}
int main()
{
int n,x,y;scanf("%ld",&n);
char enter,c;scanf("%c",&enter);
for (int i=1;i<maxn;i++)
{
f[i]=i;
h[i]=0;
deep[i]=1;
}
while (n>0)
{
n--;
scanf("%c",&c);
if (c=='M')
{
scanf("%ld%ld",&x,&y);
int fx=get(x);
int fy=get(y);
if (fx!=fy) Union(fx,fy);
}
else
{
scanf("%ld",&x);
int p=get(x);
printf("%ld\n",h[x]);
}
scanf("%c",&enter);
}
return 0;
}