BZOJ 1057 ZJOI 2007 棋盘制作 DP+悬线法

题目大意:给出一个由01形成的矩阵,问这个矩阵中最大面积的正方形和矩形,其中任意一个方块相邻的都是不同的格子。


思路:其实吧所有(i + j)&1的位置上的数字异或一下,就变成都是0或者都是1的最大正方形和矩形了。第一问就是水DP,第二问可以单调栈或者悬线。都很好写。


CODE:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 2010
using namespace std;

int m,n;
int src[MAX][MAX];
int f[MAX][MAX];

int up[MAX][MAX],_left[MAX][MAX],_right[MAX][MAX];

int main()
{
	cin >> m >> n;
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j) {
			scanf("%d",&src[i][j]);
			if((i + j)&1)
				src[i][j] ^= 1;
		}
	int ans = 0;
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			if(src[i][j]) {
				f[i][j] = min(f[i - 1][j - 1],min(f[i - 1][j],f[i][j - 1])) + 1;
				ans = max(ans,f[i][j]);
			}
	memset(f,0,sizeof(f));
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			if(!src[i][j]) {
				f[i][j] = min(f[i - 1][j - 1],min(f[i - 1][j],f[i][j - 1])) + 1;
				ans = max(ans,f[i][j]);
			}
	cout << ans * ans << endl;
	ans = 0;
	for(int i = 1; i <= m; ++i) {
		for(int j = 1; j <= n; ++j)
			_left[i][j] = src[i][j] ? _left[i][j - 1] + 1:0;
		for(int j = n; j; --j)
			_right[i][j] = src[i][j] ? _right[i][j + 1] + 1:0;
	}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			if(src[i][j] && src[i - 1][j]) {
				up[i][j] = up[i - 1][j] + 1;
				_left[i][j] = min(_left[i][j],_left[i - 1][j]);
				_right[i][j] = min(_right[i][j],_right[i - 1][j]);
				ans = max(ans,(_left[i][j] + _right[i][j] - 1) * (up[i][j] + 1));
			}
	memset(_left,0,sizeof(_left));
	memset(_right,0,sizeof(_right));
	memset(up,0,sizeof(up));
	for(int i = 1; i <= m; ++i) {
		for(int j = 1; j <= n; ++j)
			_left[i][j] = src[i][j] ? 0:_left[i][j - 1] + 1;
		for(int j = n; j; --j)
			_right[i][j] = src[i][j] ? 0:_right[i][j + 1] + 1;
	}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			if(!src[i][j] && !src[i - 1][j]) {
				up[i][j] = up[i - 1][j] + 1;
				_left[i][j] = min(_left[i][j],_left[i - 1][j]);
				_right[i][j] = min(_right[i][j],_right[i - 1][j]);
				ans = max(ans,(_left[i][j] + _right[i][j] - 1) * (up[i][j] + 1));
			}
	cout << ans << endl;
	return 0;
}


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