BZOJ 1112: [POI2008]砖块Klo

1112: [POI2008]砖块Klo

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1736  Solved: 606
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Description

N柱砖,希望有连续K柱的高度是一样的. 你可以选择以下两个动作 1:从某柱砖的顶端拿一块砖出来,丢掉不要了. 2:从仓库中拿出一块砖,放到另一柱.仓库无限大. 现在希望用最小次数的动作完成任务.

Input

第一行给出N,K. (1 ≤ k ≤ n ≤ 100000), 下面N行,每行代表这柱砖的高度.0 ≤ hi ≤ 1000000

Output

最小的动作次数

Sample Input

5 3
3
9
2
3
1

Sample Output

2

HINT

原题还要求输出结束状态时,每柱砖的高度.本题略去.

Source

 

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分析

对于一段区间，可以知道将高度改为中位数是最优的，因此我们需要做的是维护一个区间的中位数，以及小于中位数的数字之和和大于中位数的数字之和，这个可以平衡树，可以树状数组+二分查找，或者是线段树上二分，甚至是STL set，做法太多……

代码

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define ri register int

#define lim 10000000

char *c = new char[lim];

template <class T>
void read(T &x)
{
    x = 0;
    
    while (*c < '0')++c;
    
    while (*c >= '0')
        x = x*10 + *c++ - '0';
}

template <class T>
void Min(T &a, T b)
{
    if (a > b)a = b;
}

template <class T>
void Max(T &a, T b)
{
    if (a < b)a = b;
}

#define N 1000005

int n, m;
int h[N];

struct node
{
    int lt, rt, cnt;
    long long sum;
}tree[N * 6];

void build(int p, int l, int r)
{
    node &t = tree[p];
    
    t.lt = l;
    t.rt = r;
    
    t.cnt = t.sum = 0;
    
    if (l ^ r)
    {
        int mid = (l + r) >> 1;
        
        build(p << 1, l, mid);
        build(p << 1 | 1, mid + 1, r);
    }
}

void insert(int p, int pos, int val1, int val2)
{
    node &t = tree[p];
    
    t.cnt += val1;
    t.sum += val2;
    
    if (t.lt ^ t.rt)
    {
        int mid = (t.lt + t.rt) >> 1;
        
        if (pos <= mid)
            insert(p << 1, pos, val1, val2);
        else
            insert(p << 1 | 1, pos, val1, val2);
    }
}

int query1(int p, int rank)
{
    node &t = tree[p];
    
    if (t.lt == t.rt)return t.lt;
    
    if (tree[p << 1].cnt >= rank)
        return query1(p << 1, rank);
    else
        return query1(p << 1 | 1, rank - tree[p << 1].cnt);    
}

long long query2(int p, int l, int r)
{
    if (l > r)return 0LL;
    
    node &t = tree[p];
    
    if (l == t.lt && r == t.rt)
        return t.sum;
        
    int mid = (t.lt + t.rt) >> 1;
    
    if (r <= mid)
        return query2(p << 1, l, r);
    if (l > mid)
        return query2(p << 1 | 1, l, r);
    return query2(p << 1, l, mid) + query2(p << 1 | 1, mid + 1, r);
}

int query3(int p, int l, int r)
{
    if (l > r)return 0LL;
    
    node &t = tree[p];
    
    if (l == t.lt && r == t.rt)
        return t.cnt;
        
    int mid = (t.lt + t.rt) >> 1;
    
    if (r <= mid)
        return query3(p << 1, l, r);
    if (l > mid)
        return query3(p << 1 | 1, l, r);
    return query3(p << 1, l, mid) + query3(p << 1 | 1, mid + 1, r);
}

signed main(void)
{
    fread(c, 1, lim, stdin);
    
    read(n); 
    read(m);
    
    ri maxi = 0;
    ri mini = N;
    
    for (ri i = 1; i <= n; ++i)
    {
        read(h[i]);
        
        Min(mini, h[i]);
        Max(maxi, h[i]);
    }
    
    build(1, 0, N);
    
    for (ri i = 1; i < m; ++i)
        insert(1, h[i], 1, h[i]);
        
    int d = (m + 1) >> 1;
    
    long long ans = 1e18 + 9;
        
    for (ri i = m; i <= n; ++i)
    {
        insert(1, h[i], 1, h[i]);
        
        {
            int mid = d;
            
            long long q = query1(1, mid), res = 0LL;
            
            long long lc = query3(1, 0, q - 1);
            long long rc = query3(1, q + 1, N);
            
            res += lc*q - query2(1, 0, q - 1);
            res += query2(1, q + 1, N) - rc*q;
            
            Min(ans, res);
        }
        
        insert(1, h[i - m + 1], -1, -h[i - m + 1]);
    }
    
    printf("%lld
", ans);
}

@Author: YouSiki