【简●解】[HAOI2007] 理想的正方形

【题目大意】

有一个(a*b)的整数组成的矩阵，现请你从中找出一个(n*n)的正方形区域，使得该区域所有数中的最大值和最小值的差最小。

【分析】

在暴力中想到优化，模仿曾经求二维前缀和的做法，先在每行求区间长度为(n)的最大值和最小值，再在此基础上求列上的最大值和最小值，则求得的即为单个(n*n)正方形矩阵中的最大值和最小值。（仔细体味下）

用单调队列就可以搞定，和滑动窗口类似。

【Code】

一个挣扎在英语一线的苦逼(Oier)。。。敲代码背单词。。。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int N = 1000 + 5;
const int INF = 0x7fffffff;
inline int read(){
	int f = 1, x = 0;char ch;
	do { ch = getchar(); if (ch == '-') f = -1; } while (ch < '0'||ch>'9');
	do {x = x*10+ch-'0'; ch = getchar(); } while (ch >= '0' && ch <= '9'); 
	return f*x;
}
int a, b, n, aa, bb, maps[N][N], ans = INF;
int shallow[N][N];// 肤浅的 adj. ---X最大值 
int reform[N][N];// 改革，改进 v. ---X最小值 
int allergic[N][N];// 敏感的 adj. ---Y最大值
int association[N][N];//协会，交往 n. ---Y最小值 
int violence[20 * N];// 暴力行为 adj. ---队列 
int head = 1, tail = 0;

int main(){
	a = read(), b = read(), n = read();
	for (int i = 1;i <= a; ++i) {
		for (int j = 1;j <= b; ++j) {
			maps[i][j] = read();
		}
	}
	aa = a - n + 1;
	bb = b - n + 1;
	
	for (int i = 1;i <= a; ++i) {
		head = 1, tail = 0;
//		printf("rand %d : %d - %d 
", i, head, tail) ;		
		for (int j = 1;j <= b; ++j) {
			while (maps[i][violence[tail]] <= maps[i][j] && head <= tail) tail--;
			violence[++tail] = j;			
			while (head <= tail && violence[head] < j - n + 1) {
				head++;
			}
			shallow[i][j] = maps[i][violence[head]];
		}
	}
	
	for (int i = 1;i <= a; ++i) {
		head = 1, tail = 0;
//		printf("rand %d : %d - %d 
", i, head, tail) ;		
		for (int j = 1;j <= b; ++j) {
			while (maps[i][violence[tail]] >= maps[i][j] && head <= tail) tail--;
			violence[++tail] = j;			
			while (head <= tail && violence[head] < j - n + 1) {
				head++;
			}
			reform[i][j] = maps[i][violence[head]];
		}
	}

//	puts("");
//	for (int i = 1;i <= a; ++i) {
//		for (int j = 1; j <= b; ++j) {
//			printf("%d ", reform[i][j]);
//		}
//		puts("");
//	}	
	
	for (int j = 1;j <= b; ++j) {
		head = 1, tail = 0;
		for (int i = 1;i <= a; ++i) {
			while (shallow[violence[tail]][j] <= shallow[i][j] && head <= tail) tail--;
			violence[++tail] = i;			
			while (head <= tail && violence[head] < i - n + 1) {
				head++;
			}
			allergic[i][j] = shallow[violence[head]][j];
		}
	}

//	puts("");
//	for (int i = 1;i <= a; ++i) {
//		for (int j = 1; j <= b; ++j) {
//			printf("%d ", allergic[i][j]);
//		}
//		puts("");
//	}	
	
	for (int j = 1;j <= b; ++j) {
		head = 1, tail = 0;
		for (int i = 1;i <= a; ++i) {
			while (reform[violence[tail]][j] >= reform[i][j] && head <= tail) tail--;
			violence[++tail] = i;			
			while (head <= tail && violence[head] < i - n + 1) {
				head++;
			}
			association[i][j] = reform[violence[head]][j];			
		}
	}

//	puts("");
//	for (int i = 1;i <= a; ++i) {
//		for (int j = 1; j <= b; ++j) {
//			printf("%d ", association[i][j]);
//		}
//		puts("");
//	}	
	
	for (int i = n; i <= a; ++i) {
		for (int j = n; j <= b; ++j) {
			ans = min(ans, allergic[i][j] - association[i][j]);
		}
	}
	
	printf("%d", ans);
	return 0;
}