Building Block
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 151 Accepted Submission(s) : 57
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Problem Description
John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:
M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.
C X : Count the number of blocks under block X
You are request to find out the output for each C operation.
M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.
C X : Count the number of blocks under block X
You are request to find out the output for each C operation.
Input
The first line contains integer P. Then P lines follow, each of which contain an operation describe above.
Output
Output the count for each C operations in one line.
Sample Input
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
Source
2009 Multi-University Training Contest 1 - Host by TJU
题意:
M x y把x的管放在y管的上面
C x求x的下面有多少个砖块
down[i]表示i到根节点之间有多少块砖,再维护一个sum[x]数组,表示管x上一共串有多少个砖块
find每次更新down数组;
#include<iostream> #include<stdio.h> using namespace std; int par[30005]; int sum[30005];//上面有多少; int down[30005]; int findi(int x) { if(x==par[x]) return x; int p=findi(par[x]); down[x]=down[par[x]]+down[x]; return par[x]=p; } void unioni(int x,int y) { int xx=findi(x); int yy=findi(y); if(xx!=yy) { par[xx]=yy; down[xx]=sum[yy]; sum[yy]=sum[yy]+sum[xx]; } } int main() { int n; for(int i=0;i<30005;i++) sum[i]=1; for(int i=0;i<30005;i++) down[i]=0; for(int i=0;i<30005;i++) par[i]=i; scanf("%d",&n); while(n--) { char a; cin>>a; if(a=='M') { int b,c; scanf("%d%d",&b,&c); unioni(b,c); } if(a=='C') { int d; scanf("%d",&d); findi(d); printf("%d ",down[d]); } } return 0; }