POJ2985 The k-th Largest Group(treap+并查集)

题目链接

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

题目大意：给定N只猫，序号从一到N，一开始每个都是一组，然后对这群猫进行操作，可以把两只猫合成一组，然后可以查询第K大的组里有几只猫。

并查集加treap,对于每一次合并，在treap里分别将两元素删除，然后用并查集将其合并，在将合并后的结果加入treap树即可。

代码如下：

#include <iostream>#include <string.h>#include <stdlib.h>#include <algorithm>#include <stdio.h>using namespace std;const int N=200005;struct node{    int l;    int r;    int val;    int ran;    int num;    int w;}tree[N];struct query{    int c;    int a,b;}in[N];int pre[N];//并查集int s,root;int fin(int x)//并查集查找{    if(x!=pre[x])        pre[x]=fin(pre[x]);    return pre[x];}void join(int x,int y)//并查集合并函数{    int fx=fin(x),fy=fin(y);    if(fx!=fy)    {        pre[fy]=fx;    }}void lturn(int &k){    int t=tree[k].r;    tree[k].r=tree[t].l;    tree[t].l=k;    tree[t].num=tree[k].num;    tree[k].num=tree[tree[k].l].num+tree[tree[k].r].num+tree[k].w;    k=t;}void rturn(int &k){    int t=tree[k].l;    tree[k].l=tree[t].r;    tree[t].r=k;    tree[t].num=tree[k].num;    tree[k].num=tree[tree[k].l].num+tree[tree[k].r].num+tree[k].w;    k=t;}void ins(int &k,int x){    if(k==0)    {        s++;        k=s;        tree[k].val=x;        tree[k].num=tree[k].w=1;        tree[k].ran=rand();        return ;    }    tree[k].num++;    if(tree[k].val==x)        tree[k].w++;    else if(tree[k].val>x)    {        ins(tree[k].l,x);        if(tree[tree[k].l].ran<tree[k].ran)            rturn(k);    }    else    {        ins(tree[k].r,x);        if(tree[tree[k].r].ran<tree[k].ran)            lturn(k);    }}void del(int &k,int x){    if(k==0)        return;    if(tree[k].val==x)    {        if(tree[k].w>1)        {            tree[k].num--;            tree[k].w--;            return;        }        if(tree[k].l*tree[k].r==0)            k=tree[k].l+tree[k].r;        else if(tree[tree[k].l].ran<tree[tree[k].r].ran)        {            rturn(k);            del(k,x);        }        else        {            lturn(k);            del(k,x);        }    }    else if(tree[k].val<x)    {        tree[k].num--;        del(tree[k].r,x);    }    else    {        tree[k].num--;        del(tree[k].l,x);    }}int queryknum(int k,int x){    if(k==0)        return 0;    if(x<=tree[tree[k].l].num)        return queryknum(tree[k].l,x);    else if(x>tree[tree[k].l].num+tree[k].w)        return queryknum(tree[k].r,x-tree[tree[k].l].num-tree[k].w);    else        return tree[k].val;}int main(){    int data[N];    int n,m;    memset(tree,0,sizeof(tree));    memset(data,0,sizeof(data));    memset(in,0,sizeof(in));    memset(pre,0,sizeof(pre));    scanf("%d %d",&n,&m);    root=0;    s=0;    for(int i=1;i<=n;i++)    {        pre[i]=i;        data[i]=1;        ins(root,1);    }    for(int i=1;i<=m;i++)    {        scanf("%d",&in[i].c);        if(in[i].c==0)            scanf("%d %d",&in[i].a,&in[i].b);        else            scanf("%d",&in[i].a);    }    int total=n;    for(int i=1;i<=m;i++)    {        if(in[i].c==0)        {            int x=fin(in[i].a);            int y=fin(in[i].b);            if(x!=y)            {                total--;//两元素合并，树内的总元素数目减1                del(root,data[x]);                del(root,data[y]);                join(x,y);//x是y的父节点                data[x]+=data[y];//更改后的元素                ins(root,data[x]);            }        }        if(in[i].c==1)        {            printf("%d\n",queryknum(root,total-in[i].a+1));//查询为第K大的元素，所以倒过来查找，将treap插入过程的方向更改也可        }    }    return 0;}