【洛谷 P2216】 [HAOI2007]理想的正方形（二维ST表）

题目链接
做出二维(ST)表，然后(O(n^2))扫一遍就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1010;
const int MAXLOGN = 12;
int Max[MAXN][MAXN][MAXLOGN], Min[MAXN][MAXN][MAXLOGN], Log[MAXN];
int n, m, k, ans = 2147483647; 
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
int QueryMin(int x, int y){
	int p = Log[k];
	return min(min(Min[x][y][p], Min[x][y + k - (1 << p)][p]),
	           min(Min[x + k - (1 << p)][y][p], Min[x + k - (1 << p)][y + k - (1 << p)][p]));
}
int QueryMax(int x, int y){
	int p = Log[k];
	return max(max(Max[x][y][p], Max[x][y + k - (1 << p)][p]),
	           max(Max[x + k - (1 << p)][y][p], Max[x + k - (1 << p)][y + k - (1 << p)][p]));
}
int main(){
	Log[0] = -1;
	for(int i = 1; i <= 1000; ++i)
	   Log[i] = Log[i >> 1] + 1;
	memset(Max, 128, sizeof Max);
	memset(Min, 127, sizeof Min);
	n = read(); m = read(); k = read();
	for(int i = 1; i <= n; ++i)
	   for(int j = 1; j <= m; ++j)
	      Max[i][j][0] = Min[i][j][0] = read();
	for(int l = 1; l <= 10; ++l)
	   for(int i = 1; i <= n; ++i)
	      for(int j = 1; j <= m; ++j){
	         Max[i][j][l] = max(max(Max[i][j][l - 1], Max[i][min(j + (1 << (l - 1)), m)][l - 1]), 
			                    max(Max[min(i + (1 << (l - 1)), n)][j][l - 1], Max[min(i + (1 << (l - 1)), n)][min(j + (1 << (l - 1)), m)][l - 1]));
			 Min[i][j][l] = min(min(Min[i][j][l - 1], Min[i][min(j + (1 << (l - 1)), m)][l - 1]), 
			                    min(Min[min(i + (1 << (l - 1)), n)][j][l - 1], Min[min(i + (1 << (l - 1)), n)][min(j + (1 << (l - 1)), m)][l - 1]));                   
		  }
	for(int i = 1; i + k - 1 <= n; ++i)
		for(int j = 1; j + k - 1 <= m; ++j)
		   ans = min(ans, QueryMax(i, j) - QueryMin(i, j));
	printf("%d
", ans);
	return 0;
}