poj2985线段树求第k大的数(并查集)

The k-th Largest Group

Time Limit: 2000MS Memory Limit: 131072KTotal Submissions: 7255 Accepted: 2335

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

10 100 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

//线段树每个节点维护的是group大小在这个区间的个数//每次更新时，包含a[x]和a[y] group大小的区间分别减1//而包含a[x]+a[y]的区间加1，询问的时候，先询问右子树//如果右子树的cnt>=k，则去右子树继续查询，否则去左子树//并且k-=rson的cnt。#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 200005#define ls(o) o<<1#define rs(o) (o<<1)|1#define Mid(a,b) (a + b)>>1#define Fill(s,e,xx) for(int i=s;i<e;i++)a[i]=xx;int fa[maxn],Rank[maxn],n,m;void init(int n)//一定要记得初始化{    for(int i=0;i<=n;i++)fa[i]=i,Rank[i]=0;}int find_fa(int x)//查询是否已经在同一组了{    if(fa[x]!=x)fa[x]=find_fa(fa[x]);    return fa[x];}void Union(int x,int y)//在这里并不需要这么合并，^-^，算是无用吧{    x = find_fa(x);    y = find_fa(x);    if(x==y)return;    if(Rank[x]<Rank[y]){        fa[x] = y;    }    else{        fa[y] = x;        if(Rank[x]==Rank[y])Rank[x]++;    }}struct line_tree{    int l,r;//存储区间信息    int cnt;//存储在该区间内，有多少区间在该区间内的group}tree[maxn<<2];void build(int tl,int tr,int o){    tree[o].l=tl,tree[o].r=tr;    if(tl==1)tree[o].cnt = n;    else tree[o].cnt = 0;    if(tl == tr){        return;    }    int mid = Mid(tl,tr);    build(tl,mid,ls(o));    build(mid+1,tr,rs(o));}void Insert(int x,int o,int c){    tree[o].cnt+=c;    if(tree[o].l == tree[o].r)return;    if(x<=tree[ls(o)].r)Insert(x,ls(o),c);    else Insert(x,rs(o),c);}int query(int k,int o=1){    if(tree[o].l==tree[o].r){        printf("%d\n",tree[o].l);        return 0;    }    if(k<=tree[rs(o)].cnt)query(k,rs(o));//右子树有大于k个数，第k个数肯定在右边。。。。//求的是第k大数。。。。所以从右子树查询    else query(k-tree[rs(o)].cnt,ls(o));}int a[maxn];//存储group信息int  main(){    while(scanf("%d%d",&n,&m)==2){        Fill(0,n+1,1);        build(1,n,1);init(n);        for(int i=0;i<m;i++){            int com;            scanf("%d",&com);            if(com==0){                int x,y;                scanf("%d%d",&x,&y);                x = find_fa(x),y=find_fa(y);                if(x==y)continue;                Insert(a[x],1,-1);                Insert(a[y],1,-1);                Insert(a[x]+a[y],1,1);                fa[y] = x;//两个一定要一致。。。。。。。。。。                a[x]+=a[y];            }            else{                int k;                scanf("%d",&k);                query(k);            }        }    }    return 0;}