2761 Feed the dogs 求任一区间的第k小数 线段树+划分树

Feed the dogs

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8160		Accepted: 2218

Description

Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.

Your task is to help Jiajia calculate which dog ate the food after each feeding.

Input

The first line contains n and m, indicates the number of dogs and the number of feedings.

The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.

Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.

You can assume that n<100001 and m<50001.

Output

Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

Sample Input

7 2
1 5 2 6 3 7 4
1 5 3
2 7 1

Sample Output

3
2

http://blog.sina.com.cn/s/blog_5f5353cc0100ki2e.html

#include
#include
#include
#include
using namespace std;
const int N=101000;
const int DEEP=20;//划分树最多层数
struct Node
{
    int l,r;
};
Node tree[N*4];//线段树
int data[N];//数据
int seg[DEEP][N];//划分树
int LessMid[DEEP][N];//表示在[L，R]内有几个数小于等于date[mid]

void buildtree(int root,int l,int r,int d)//节点，左右区间，层数
{
    tree[root].l=l,tree[root].r=r;
    if(l==r) return ;//叶子节点
    int mid=(l+r)>>1;
    int lsame=mid-l+1;//左面最多可放几个与data[mid]相同的数
    for(int i=l;i<=r;i++)//得出实际能放几个相同的数
    {
        if(seg[d][i]    }

    int tl=l,tr=mid+1,same=0;//划分树 ,tl,tr表示数的左右子树的起点
    for(int i=l;i<=r;i++)
    {
        if(i==l) LessMid[d][i]=0;//表示在[L，R]内有几个数小于等于date[mid]
        else LessMid[d][i]=LessMid[d][i-1];

        if(seg[d][i]        else if(seg[d][i]>data[mid]) seg[d+1][tr++]=seg[d][i];//划分到右子树
        else //相等情况
        {
            if(same            else seg[d+1][tr++]=seg[d][i];
        }
    }
    buildtree(root<<1,l,mid,d+1);
    buildtree((root<<1)+1,mid+1,r,d+1);
}
//查询[l,r]中的第k小数
int Query(int root,int l,int r,int d,int cnt)//节点，要查询的区间,层数，第cnt小数
{
    if(l==r) return seg[d][l];

    int s;//表示在[l,r]中有几个小于等于data[mid]的个数
    int ss;//表示在[tree[root].l,l-1]中有几个小于等于data[mid]的个数
    if(l==tree[root].l) s=LessMid[d][r],ss=0;
    else s=LessMid[d][r]-LessMid[d][l-1],ss=LessMid[d][l-1];
    if(s>=cnt) return Query(root<<1,tree[root].l+ss,tree[root].l+ss+s-1,d+1,cnt);
    else
    {
        int mid=(tree[root].l+tree[root].r)>>1;
        int bb=l-tree[root].l-ss;//表示[tree[root].l,l-1]中有多少个分到右面
        int b=r-l+1-s;//表示[l,r]有多少个分到右面
        return Query((root<<1)+1,mid+bb+1,mid+bb+b,d+1,cnt-s);
    }
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&data[i]);
            seg[1][i]=data[i];//划分树第一层
        }
        sort(data+1,data+1+n);
        buildtree(1,1,n,1);
        while(m--)
        {
            int l,r,k;scanf("%d%d%d",&l,&r,&k);
            int res=Query(1,l,r,1,k);//查询[l,r]中的第k小数
            printf("%d/n",res);
        }
    }
    return 0;
}