POJ1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 24665		Accepted: 8653
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

//同P2342 叠积木
#include<cstdio>
using namespace std;
const int N=3e5+5;
int n,fa[N],top[N],cnt[N];
int find(int x){
    if(fa[x]==x) return x;
    int t=fa[x];
    fa[x]=find(fa[x]);
    fa[x]=fa[t];
    top[x]=top[t];
    cnt[x]=cnt[t]+cnt[x];
    return fa[x];
}
int main(){
    scanf("%d",&n);char s[10];
    for(int i=1;i<=30000;i++) fa[i]=top[i]=i;
    for(int i=1,a,b,x,y;i<=n;i++){
        scanf("%s",s);
        if(s[0]=='M'){
            scanf("%d%d",&a,&b);
            x=find(a);y=find(b);
            fa[x]=y;find(top[y]);
            cnt[x]=cnt[top[y]]+1;
            top[y]=top[x];
        }
        else{
            scanf("%d",&x);find(x);
            printf("%d\n",cnt[x]);
        }
    }
    return 0;
}

 

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6478001.html
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