poj 3494 Largest Submatrix of All 1’s

       最近在学习单调桟，这个题目听有意思的，需要稍微动动心思，往poj 2559上想。但是时间总是1700MS左右，真不知道网上那些100ms是怎么出来的。

/**
 *  poj_3494.cpp
 *
 */
#include 
#include 
#include 
using namespace std;

#define	MN	2010
#define hpmax(a,b)		((a)>(b)?(a):(b))
#define hpmin(a,b)		((a)<(b)?(a):(b))

char	matrix[MN][MN];
int	bound[MN][MN], lmin[MN], rmin[MN];
int stack[MN];
int m, n, p;
int main() 
{
	int i, j, ans, tmp, left, right;

	while ( scanf(" %d%d", &m, &n ) != EOF ) {
		for ( i = 1; i <= m; ++ i ) {
			for ( j = 1; j <= n; ++ j )
				scanf(" %c", &matrix[i][j] );
		}
		for ( i = 1; i <= m; ++ i ) {
			tmp = 0;
			for ( j = n; j >= 1; -- j ) {
				if ( '1' == matrix[i][j] ) 
					++ tmp;
				else 	
					tmp = 0;
				bound[i][j] = tmp;
			}
		} 
		ans = 0;
		for ( j = 1; j <= n; ++ j ) {

			stack[0] = 0;
			bound[0][j] = -1;
			stack[1] = 1;
			lmin[1] = 0;
			p = 1;
			for ( i = 2; i <= m; ++ i ) {
				while ( p > 0 && bound[ stack[p] ][j] > bound[i][j] ) {
					rmin[ stack[p--] ] = i;
				}
				if ( bound[stack[p]][j] == bound[i][j])
					lmin[i] = lmin[ stack[p] ];
				else 
					lmin[i] = stack[p];
				stack[++p] = i;
			}
			while ( p > 0 ) {
				rmin[ stack[p--] ] = m+1;
			}

			for ( i = 1; i <= m; ++ i ) {
				tmp = bound[i][j] * ( rmin[i] - lmin[i] - 1 );
				if ( tmp > ans )
					ans = tmp;
			}	
		}
		printf("%d\n", ans );
	}
	return 0;
}