POJ 题目2985 The k-th Largest Group（线段树单点更新求第k大值，并查集）

The k-th Largest Group

Time Limit: 2000MS Memory Limit: 131072KTotal Submissions: 7869 Accepted: 2534

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 catj is in, thus creating a new group. On top of that, Newman wants to know the size of thek-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. IfC = 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); IfC = 1, then there is only one number k (1 ≤ k ≤ the current number of groups) following indicating Newman wants to know the size of thek-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

10 100 1 21 40 3 41 20 5 61 10 7 81 10 9 101 1

Sample Output

12222

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

POJ Monthly--2006.08.27, zcgzcgzcg

题目大意：n只猫，编号1~n，m个操作，0 i j意思是把i在的组并到j，1 k的意思是求第k大组的大小

注释的是我的思路，这个写法是参照别人的思路，他的比我的稍微快个100来ms，

ac代码

#include<stdio.h>#include<string.h>int pre[200020],num[200020];int node[200020<<2],n,m;void build_tr(int l,int r,int tr){if(l==1)node[tr]=n;elsenode[tr]=0;if(l==r){/*if(l==1)node[tr]=n;elsenode[tr]=0;*/return;}int mid=(l+r)>>1;build_tr(l,mid,tr<<1);build_tr(mid+1,r,tr<<1|1);//node[tr]=node[tr<<1]+node[tr<<1|1];}void init(int n){int i;for(i=1;i<=n;i++){num[i]=1;pre[i]=i;}}int find(int x){if(x==pre[x])return x;return pre[x]=find(pre[x]);}void update(int pos,int add,int l,int r,int tr){node[tr]+=add;if(l==r){//node[tr]+=add;return;}int mid=(l+r)>>1;if(pos<=mid){update(pos,add,l,mid,tr<<1);}elseupdate(pos,add,mid+1,r,tr<<1|1);//node[tr]=node[tr<<1]+node[tr<<1|1];}int query(int k,int l,int r,int tr){if(l==r){return l;}int mid=(l+r)>>1;if(node[tr<<1|1]>=k)query(k,mid+1,r,tr<<1|1);elsequery(k-node[tr<<1|1],l,mid,tr<<1);}int main(){//int n,m;while(scanf("%d%d",&n,&m)!=EOF){build_tr(1,n,1);init(n);int i;for(i=0;i<m;i++){int op;scanf("%d",&op);if(!op){int a,b;scanf("%d%d",&a,&b);int fa=find(a),fb=find(b);if(fa!=fb){pre[fa]=fb;update(num[fa],-1,1,n,1);update(num[fb],-1,1,n,1);update(num[fa]+num[fb],1,1,n,1);num[fb]+=num[fa];}}else{int a;scanf("%d",&a);printf("%d\n",query(a,1,n,1));}}}}