POJ 2761 Feed the dogs

求区间第k小，划分树，又一种神奇的数据结构。2104也可以用这个代码AC，那道题是裸的求第k小。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<algorithm>

using namespace std;

const int MAXN = 100010;

struct Node

{

    int l, r;

}T[MAXN << 2];



int d[MAXN], s[MAXN], t[21][MAXN], tol[21][MAXN];



void build(int level, int rt, int l, int r)

{

    int i;

    T[rt].l = l, T[rt].r = r;

    if(l == r) return;

    int mid = l + r >> 1;

    int lsame = mid - l + 1;

    for(i = l; i <= r; i ++)

    {

        if(t[level][i] < s[mid])

            lsame --;

    }

    int Ll = l, Lr = mid, Rl = mid + 1, Rr = r;

    int Ltot = 0, Rtot = 0;

    for(i = l; i <= r; i ++)

    {

        if(i == l)

            tol[level][i] = 0;

        else

            tol[level][i] = tol[level][i - 1];

        if(t[level][i] < s[mid])

        {

            tol[level][i] ++;

            t[level + 1][Ll + Ltot ++] = t[level][i];

        }

        else if(t[level][i] > s[mid])

        {

            t[level + 1][Rl + Rtot ++] = t[level][i];

        }

        else

        {

            if(lsame > 0)

            {

                lsame --;

                t[level + 1][Ll + Ltot ++] = t[level][i];

                tol[level][i] ++;

            }

            else

            {

                t[level + 1][Rl + Rtot ++] = t[level][i];

            }

        }

    }

    build(level + 1, rt << 1, Ll, Lr);

    build(level + 1, rt << 1 | 1, Rl, Rr);

}



int query(int level, int rt, int l, int r, int k)

{

    if(l == r) return t[level][l];

    int nl, nls;

    if(l == T[rt].l)

        nls = 0;

    else

        nls = tol[level][l - 1];

    nl = tol[level][r] - nls;

    if(nl >= k)

    {

        return query(level + 1, rt << 1, T[rt].l + nls,

            T[rt].l + nl + nls - 1, k);

    }

    else

    {

        int mid = (T[rt].l + T[rt].r) >> 1;

        int nre = l - T[rt].l + 1 - nls;

        int nr = r - l + 1 - nl;

        return query(level + 1, rt << 1 | 1, mid + nre,

            mid + nr + nre - 1, k - nl);

    }

}



int main()

{

    int n, m, i;

    while(scanf("%d%d", &n, &m) == 2)

    {

        for(i = 1; i <= n; i ++)

        {

            scanf("%d", &d[i]);

            t[1][i] = s[i] = d[i];

        }

        sort(s + 1, s + n + 1);

        build(1, 1, 1, n);

        int l, r, k;

        while(m --)

        {

            scanf("%d%d%d", &l, &r, &k);

            printf("%d\n", query(1, 1, l, r, k));

        }

    }

    return 0;

}