BZOJ 1057 ZJOI 2007 棋盘制作 DP+悬线法

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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 2010using 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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