poj 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 22163 Accepted: 7777Case 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

题意：有两种操作M 把X堆得碟子放到Y堆上 C X下面有几个碟子

带权并查集，以x为根结点的所有元素减去以X到根结点的距离再减一

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int i,pre[30100];int vis[30100];//x到根节点的距离 int dis[30100];//以X为根节点元素的个数 void init(){for(i=1;i<=30100;i++){pre[i]=i;vis[i]=0;dis[i]=1;}}int find(int x){if(x==pre[x])return x;else{int root=find(pre[x]);vis[x]=vis[x]+vis[pre[x]];return pre[x]=root;} }  void merge(int x,int y) { int fx=find(x); int fy=find(y); pre[fy]=fx; vis[fy]=dis[fx]; dis[fx]+=dis[fy]; } int main() { int t,a,b; char str[10]; while(~scanf("%d",&t)) { init(); while(t--) { scanf("%s",str); if(str[0]=='M') { scanf("%d%d",&a,&b); merge(a,b); } else { scanf("%d",&a); int fx=find(a); printf("%d\n",dis[fx]-vis[a]-1); } } } return 0; }