Cube Stacking（并查集）

http://poj.org/problem?id=1988

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 22312 Accepted: 7823Case 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

6M 1 6C 1M 2 4M 2 6C 3C 4

Sample Output

102

Source

USACO 2004 U S Open

简单并查集，入门好题

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int SIZE=3e4+10;int parent[SIZE];int under[SIZE],sum[SIZE];int getroot(int x){    if(parent[x]==x)return x;    int fx=getroot(parent[x]);//递归获得根，前面的层数的返回最后的根节点    under[x]+=under[parent[x]];//是parent[x]不是fx，给的例子中先最新的fx的under是0，而加的应该是原先的under值,    return parent[x]=fx;}//一次性把一条路径上的点更新了，每个parent[x]都是保存了当初一次的under的值，不太好表达void Merge(int x,int y){    int fx=getroot(x),fy=getroot(y);    if(fx==fy)return ;    parent[fx]=fy;    under[fx]=sum[fy];    sum[fy]+=sum[fx];}int main(){    int p,x,y,c;    while(scanf("%d",&p)!=EOF){        char s[10];        for(int i=1;i<SIZE;i++){            under[i]=0;            sum[i]=1;            parent[i]=i;        }        while(p--){            scanf("%s",s);            if(s[0]=='M'){                scanf("%d%d",&x,&y);                Merge(x,y);            }            else {                scanf("%d",&c);                getroot(c);                printf("%d\n",under[c]);            }        }        return 0;    }}/**M 1 2M 1 3M 1 4画画图***/