hdu 4006 线段树插入与单点更新

最新推荐文章于 2021-03-14 00:15:15 发布

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最新推荐文章于 2021-03-14 00:15:15 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/yu_ming_cong/article/details/47363253

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本文探讨了小明和小宝玩数字游戏的过程，详细解释了如何通过选择性地写数字或询问第k大的数字来解决问题。文章提供了一个利用线段树的数据结构来高效查询第k大数字的方法，并附上了代码实现。

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The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)

Problem Description

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

Input

There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.

Output

The output consists of one integer representing the largest number of islands that all lie on one line.

Sample Input

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

Sample Output

1
2
3

Hint
Xiao  Ming  won't  ask  Xiao  Bao  the  kth  great  number  when  the  number  of  the  written number is smaller than k. (1=<k<=n<=1000000).

线段树每个节点保存当前编号的数出现的次数。每次查询从小到大第N-K+1个数（N为当前数的总数）。

代码：

#include <iostream>
#define MAXN 1000100
#include <stdio.h>
using namespace std;
struct Tree
{
    int l,r;
    int res;
} tree[MAXN*4];
int ans[MAXN],n,k;
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    if(l==r)
    {
        tree[root].res=0;
        return;
    }
    int mid=(r+l)/2;
    build(root*2,l,mid);
    build(root*2+1,mid+1,r);
    tree[root].res=tree[root<<1].res+tree[root<<1|1].res;
}
int query(int root,int pos,int l,int r)
{  int ret;
    if(l==r)
    {
        return l;
    }

    else
    {int m=(l+r)/2;
        if(pos<=tree[root*2].res)
            ret=query(root*2,pos,l,m);
        else
            ret=query(root*2+1,pos-tree[root*2].res,m+1,r);
    }
    return ret;
}
void update(int root,int l,int r,int pos)
{
    if(l==r)
    {
        tree[root].res++;
        return;
    }
    int m=(l+r)/2;
    if(pos<=m)
    update(root<<1,l,m,pos);
    else
    update(root<<1|1,m+1,r,pos);
    tree[root].res=tree[root<<1].res+tree[root<<1|1].res;
}
void solve()
{
    char a[5];
    int ff;
    build(1,1,MAXN);
    int counts=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%s",a);
        if(a[0]=='I')
        {
        scanf("%d",&ff);
        update(1,1,MAXN,ff);
        counts++;
        }
        else
        printf("%d\n",query(1,counts-k+1,1,MAXN));
    }

}
int main()
{    while(scanf("%d%d",&n,&k)!=EOF)
        solve();
    return 0;
}