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划分树 + 二分

二分枚举第k小的数与h比较大小。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
const int MAXN = 100100;
struct Node
{
    int l, r;
}T[MAXN << 2];

int d[MAXN], s[MAXN], t[35][MAXN], tol[35][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 BS(int r, int h, int L, int R)
{
    int l = 0, mid;
    while(l < r)
    {
        mid = (l + r) >> 1;
        if(query(1, 1, L, R, mid) > h)
            r = mid ;
        else l = mid + 1;
    }
    return l;
}

int main()
{
    int n, m, i, T, cas;
    scanf("%d", &T);
    for(cas = 1; cas <= T; cas ++)
    {
        scanf("%d%d", &n, &m);
        for(i = 1; i <= n; i ++)
        {
            //d[i] = 1;
            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, h;
        printf("Case %d:\n", cas);
        while(m --)
        {
            scanf("%d%d%d", &l, &r, &h);
            l ++, r ++;
            if(query(1, 1, l, r, r - l + 1) <= h)
                printf("%d\n", r - l + 1);
            else if(query(1, 1, l, r, 1) > h)
                puts("0");
            else{
                int tt = BS(r - l + 1, h, l, r);
                //printf("%d\n", tt - 1);

                while(query(1, 1, l, r, tt) > h)
                {
                    tt --;
                }
                printf("%d\n", tt);

            }
        }
    }
    return 0;
}
posted on 2012-09-23 17:20  找回失去的  阅读(1065)  评论(0编辑  收藏  举报