HDU 2818 Building Block 带权并查集
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Building Block
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5966 Accepted Submission(s): 1853
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.
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:
有n块砖和p次操作:
M x y:表示将编号为x所在的这一列砖移到编号为y在的那一列砖的上面,如果两块砖已经在同一列了,就跳过。
C x:表示询问编号为x的砖的下面有多少砖。
坑点:砖的编号实际为0~30000 - 1
分析:
还是关于一些集合合并的问题,只不过这次要询问下面有多少块砖。所以只是单纯的并查集是不够的。
我们用rk[x]表示编号为x的砖下面有多少块砖。将在上面的砖连到它下面的砖上。则路径压缩时合并:rk[x]+=rk[fa[x]];
对于需要合并的砖a、b:
不难发现,对于移动到上方的砖a,现在a所在的那一列的每一块砖的rk都需要加上siz[b],
//siz[i]表示i所在的这一列有多少块砖
可以直接在root a上加上siz[b],注意合并后siz的改变。
至于查询,可以再查找一下它现在的根,如果之前有根的改变,那么在查找的过程中就会将改变的值加上。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();} return x*f;}const int N = 30000 + 10;int p;int fa[N],rk[N],siz[N];int find(int x){ if(x==fa[x]) return fa[x]; int d=find(fa[x]); rk[x]+=rk[fa[x]]; return fa[x]=d;}void link(int x,int y){ int fx=find(x),fy=find(y); if(fx==fy) return ; rk[fx]+=siz[fy]; siz[fy]+=siz[fx]; fa[fx]=fy;}int main(){ p=read(); for(register int i=0;i<=N-10;i++) siz[i]=1,fa[i]=i,rk[i]=0; for(register int j=1;j<=p;j++){ char s[5];scanf("%s",s); if(s[0]=='M'){ int x=read(),y=read(); link(x,y); } else{ int x=read();find(x); printf("%d\n",rk[x]); } } return 0;}
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