poj 2104 K-th Number

類型：劃分樹【學習資料】

題目：http://poj.org/problem?id=2104

來源：Northeastern Europe 2004, Northern Subregion

思路    代碼源自：http://blog.sina.com.cn/s/blog_5f5353cc0100ki2e.html

 http://blog.csdn.net/zxy_snow/article/details/6681086【資料】

// poj 2104 K-th Number
// ac 16344K 907MS
#include<cstdio>
#include<algorithm>
using namespace std;

#define FORE(i, a, b) for(i = a; i <= b; ++i)
const int N = 100000 + 10;

struct NODE {
    int l,r;
}tr[N * 4];
int data[N], seg[20][N], lessMid[20][N];

void Build(int l, int r, int root, int d) {
    tr[root].l = l, tr[root].r = r;
    if( l == r )
        return;
    int i, mid = (l + r) >> 1;
    int lsame = mid - l + 1;   //這個變量表示左邊最多可放幾個與data[mid]相同的數
    FORE(i, l, r) //得出實際能放幾個相同的數
        if( seg[d][i] < data[mid] )
            --lsame;
    int tl = l, tr = mid + 1, same = 0;
    FORE(i, l, r) {
        if( i == l )
            lessMid[d][i] = 0;  //表示在[L，R]內有幾個數小於等於data[mid]
        else
            lessMid[d][i] = lessMid[d][i - 1];
        if( seg[d][i] < data[mid] ) {
            ++lessMid[d][i];
            seg[d + 1][tl++] = seg[d][i];
        }
        else if( seg[d][i] > data[mid] )
            seg[d + 1][tr++] = seg[d][i];
        else {
            if( same < lsame ) {
                ++same, ++lessMid[d][i];
                seg[d + 1][tl++] = seg[d][i];
            }
            else
                seg[d + 1][tr++] = seg[d][i];
        }
    }
    Build(l, mid, root * 2, d + 1);
    Build(mid + 1, r, root * 2 + 1, d + 1);
}

int Query(int l, int r, int root, int d, int cnt) {
    if( l == r )
        return seg[d][l];
    int s;   //表示在[l，r]內有幾個小於等於data[mid]的個數
    int ss;  //表示在[tr[root].l,l-1]內有幾個小於等於data[mid]的個數
    if( l == tr[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(tr[root].l + ss, tr[root].l + ss + s - 1, root * 2, d + 1, cnt);
    else {
        int mid = (tr[root].l + tr[root].r) >> 1;
        int bb = l - tr[root].l - ss;  //表示 [tr[root].l , l-1 ]有多少個分到右邊
        int b = r - l + 1 - s;           //表示 [l , r]有多少個分到右邊
        return Query(mid + bb + 1, mid + bb + b, 2 * root + 1, d + 1 ,cnt - s);
    }
}

int main() {
    int n, m, l, r, cnt, i;
    while(scanf("%d%d", &n, &m) != EOF) {
        FORE(i, 1, n) {
            scanf("%d", data + i);
            seg[1][i] = data[i];
        }
        sort(data + 1,data + n + 1);
        Build(1, n, 1, 1);
        while( m-- ) {
            scanf("%d%d%d", &l, &r, &cnt);
            printf("%d\n", Query(l, r, 1, 1, cnt));
        }
    }
    return 0;
}