poj1988_Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 24130		Accepted: 8468
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

唉，一个并查集的题，改了好久。。。终于a了。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int up[300005];//该数上面的数
int n[300005];//该集合总数
int a[300005];//保存他的父节点

int father(int t){
    if(t==a[t]){
        return t;
    }
    int fa=a[t];
    a[t]=father(a[t]);
    up[t]+=up[fa];

    return a[t];
}

int main()
{
    int p;
    //char c;
    int t1,t2;
    int t3;
while(scanf("%d",&p)!=EOF){
    for(int i=0;i<=p;i++){
        a[i]=i;
        up[i]=0;
        n[i]=1;
    }
    char s[10];
    for(int i=0;i<p;i++){
        scanf("%s",s);
        if(s[0]=='M'){
            scanf("%d%d",&t1,&t2);
            int fa=father(t1);
            int fb=father(t2);
            if(fa!=fb){
                a[fb]=fa;
                up[fb]+=n[fa];
                n[fa]+=n[fb];
            }
        }else{
            scanf("%d",&t3);
            int ans=father(t3);
            printf("%d
",n[ans]-up[t3]-1);
        }
    }
}
    return 0;
}