标签：char const ges using combine bin stream pos cst
| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 8807 | Accepted: 2875 |
Description
Newman likes playing with cats. He possesses lots of cats in his home. Because the number of cats is really huge, Newman wants to group some of the cats. To do that, he first offers a number to each of the cat (1, 2, 3, …, n). Then he occasionally combines the group cat i is in and the group cat j is in, thus creating a new group. On top of that, Newman wants to know the size of the k-th biggest group at any time. So, being a friend of Newman, can you help him?
Input
1st line: Two numbers N and M (1 ≤ N, M ≤ 200,000), namely the number of cats and the number of operations.
2nd to (m + 1)-th line: In each line, there is number C specifying the kind of operation Newman wants to do. If C = 0, then there are two numbers i and j (1 ≤ i, j ≤ n) following indicating Newman wants to combine the group containing the two cats (in case these two cats are in the same group, just do nothing); If C = 1, then there is only one number k (1 ≤ k ≤ the current number of groups) following indicating Newman wants to know the size of the k-th largest group.
Output
For every operation “1” in the input, output one number per line, specifying the size of the kth largest group.
Sample Input
10 10 0 1 2 1 4 0 3 4 1 2 0 5 6 1 1 0 7 8 1 1 0 9 10 1 1
Sample Output
1 2 2 2 2
Hint
When there are three numbers 2 and 2 and 1, the 2nd largest number is 2 and the 3rd largest number is 1.
Source
并查集维护连通分量大小，树状数组求cc中第k大值
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const int N=2e5+5; inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} return x*f; } int n,m,op,x,y,k; int fa[N],size[N],tot=0; inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);} int c[N]; inline int lowbit(int x){return x&-x;} inline void add(int p,int v){ for(;p<=n;p+=lowbit(p)) c[p]+=v; } inline int sum(int p){ int res=0; for(;p>0;p-=lowbit(p)) res+=c[p]; return res; } inline int kth(int k){ int x=0,cnt=0; for(int i=16;i>=0;i--){ x+=(1<<i); if(x>=n||cnt+c[x]>=k) x-=(1<<i); else cnt+=c[x]; } return x+1; } int main(){ n=read();m=read(); for(int i=1;i<=n;i++) fa[i]=i,size[i]=1,tot++; add(1,n); for(int i=1;i<=m;i++){ op=read(); if(!op){ x=read();y=read(); int f1=find(x),f2=find(y); if(f1!=f2){ fa[f1]=f2; add(size[f1],-1); add(size[f2],-1); size[f2]+=size[f1]; add(size[f2],1); tot--; } // printf("%d %d %d %d\n",f1,f2,size[f1],size[f2]); }else{ k=tot-read()+1;//printf("k %d\n",k); printf("%d\n",kth(k)); } } }
POJ2985 The k-th Largest Group[树状数组求第k大值 并查集]
标签：char const ges using combine bin stream pos cst
原文地址：http://www.cnblogs.com/candy99/p/6067401.html
