[HAOI2007]理想的正方形

嘟嘟嘟

 

这题一种做法是单调队列，像二维的滑动窗口。或是二维st表。我用的是第二种方法。

 

首先预处理的时候，对于一个边长为2^k的正方形的最大值（最小值同理），应该由位于这个正方形的四个角上的边长为2^k-1的小正方形转化而来。令Max_i,j,k表示左上角为(i, j) 的边长为2k的正方形的最大值，则

Max_i,j,k = max(Max_{i,j,k - 1}, Max_{i,j + (1 << (k - 1)),k - 1}, Max_{i + (1 << (k - 1)),j,k - 1}, Max_{i + (1 << (k - 1)),j + (1 << (k - 1)),k - 1});

 

查询的时候，我们先求一个值k_n，表示在2^k <= n的前提下k的最大值。求完这个就像一维st表那样，只不过从这个n * n的正方形的四个角所在的小正方形转化而来，如下图

 

这个n * n的正方形的最大值就由这四个小正方形最大值转化而来。

这样转移方程就出来了

maxx = max(Max_i,j,kn, Max_{i,j + n - (1 << kn),kn}, Max_{i + n - (1 << kn),j,kn}, Max_{i + n - (1 << kn),j + n - (1 << kn),kn})

有一个小优化就是预处理的时候k只用枚举到kn，而不是到(1 << k) == min(a, b).

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<stack>
#include<queue>
#include<vector>
#include<cctype>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e3 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int x, y, n, a[maxn][maxn];

int Max[maxn][maxn][7], Min[maxn][maxn][7], nk;
void rmq()
{
    for(int i = 1; i <= x; ++i)
        for(int j = 1; j <= y; ++j) Max[i][j][0] = Min[i][j][0] = a[i][j];
    for(int k = 1; (1 << k) <= n; ++k)
        for(int i = 1; i + (1 << k) - 1 <= x; ++i)
            for(int j = 1; j + (1 << k) - 1 <= y; ++j)
            {
                Max[i][j][k] = max(max(Max[i][j][k - 1], Max[i][j + (1 << (k - 1))][k - 1]), max(Max[i + (1 << (k - 1))][j][k - 1], Max[i + (1 << (k - 1))][j + (1 << (k - 1))][k - 1]));
                Min[i][j][k] = min(min(Min[i][j][k - 1], Min[i][j + (1 << (k - 1))][k - 1]), min(Min[i + (1 << (k - 1))][j][k - 1], Min[i + (1 << (k - 1))][j + (1 << (k - 1))][k - 1]));
            }
    while((1 << (nk + 1)) <= n) nk++;
}

int ans = 2147483647;

int main()
{
    x = read(); y = read(); n = read();
    for(int i = 1; i <= x; ++i)
        for(int j = 1; j <= y; ++j) a[i][j] = read();
    rmq();
    for(int i = 1; i <= x - n + 1; ++i)
        for(int j = 1; j <= y - n + 1; ++j)
        {
            int _max = max(max(Max[i][j][nk], Max[i][j + n - (1 << nk)][nk]), max(Max[i + n - (1 << nk)][j][nk], Max[i + n - (1 << nk)][j + n - (1 << nk)][nk]));
            int _min = min(min(Min[i][j][nk], Min[i][j + n - (1 << nk)][nk]), min(Min[i + n - (1 << nk)][j][nk], Min[i + n - (1 << nk)][j + n - (1 << nk)][nk]));
            ans = min(ans, _max - _min);
        }
    write(ans); enter;
    return 0;
}