POJ 1988 Cube Stacking(带权并查集)

题目链接

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 24006 Accepted: 8431Case 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

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int MAXN=30000+100;int pa[MAXN],under[MAXN],sum[MAXN];void init(){for(int i=1;i<MAXN;i++) pa[i]=i,sum[i]=1;}int find(int x){if(pa[x]==x) return x;int t=find(pa[x]);under[x]+=under[pa[x]];pa[x]=t;return pa[x];}int main(){int n;scanf("%d",&n);init();while(n--){char s[10];scanf("%s",s);if(s[0]=='M'){int x,y;scanf("%d%d",&x,&y);int fx=find(x),fy=find(y);if(fx!=fy){pa[fx]=fy;    under[fx]+=sum[fy];    sum[fy]+=sum[fx];}}else{int p;scanf("%d",&p);find(p);printf("%d\n",under[p]);}}return 0;}