POJ1988 Cube Stacking

 

Time Limit: 2000MS

Memory Limit: 30000KB

64bit IO Format: %lld & %llu

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

 

偏移量并查集。

维护每个块所在栈的块数，每个块下面的块数。

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=100010;
int fa[mxn],und[mxn],cnt[mxn];
int find(int x){
    if(fa[x]==x)return x;
    int ff=fa[x];
    fa[x]=find(fa[x]);
    und[x]+=und[ff];
    return fa[x];
}
int n,m;
int main(){
    scanf("%d",&m);
    char op[3];int x;
    for(x=1;x<=30000;x++)fa[x]=x,cnt[x]=1,und[x]=0;
    int u,v;
    while(m--){
        scanf("%s",op);
        if(op[0]=='M'){
            scanf("%d%d",&u,&v);
            u=find(u);
            v=find(v);
            if(u!=v){
                fa[u]=v;
                und[u]+=cnt[v];
                cnt[v]+=cnt[u];
                cnt[u]=0;
            }
        }
        else{
            scanf("%d",&u);
            find(u);
            printf("%d\n",und[u]);
        }
    }
    return 0;
}