pku3494_Largest Submatrix of All 1’s

http://poj.org/problem?id=3494

像hdu2830那样处理矩阵，每一行保存的值是第 i 列以a[i]结尾的连续1的高度。。
（ http://www.cppblog.com/A-wu/archive/2010/11/29/134980.html ）

然后每一行就像pku2559和pku2796那样处理，得到每一行的最大区间值；再通过枚举 m 行得到整个矩阵的最大值。。

注意：这道题的l[]和r[]存放的不是左右边界的下标，而是h[i]到其左右边界的个数。

01 #include <cstdio>
02 #include <cstring>
03 #include <algorithm>
04 using namespace std;
05
06 const int N = 2010;
07 int stack [N ], l [N ], r [N ], g [N ][N ];
08 int n , m;
09
10 int work( int h [])
11 {
12         //求左边界
13     int top = 0;
14     stack [ 0 ] = 0;
15     h [ 0 ] = - 1;
16     for ( int i = 1; i <= n; i ++)
17     {
18         while ( h [ stack [ top ]] >= h [ i ])
19             top --;
20         l [ i ] = i - stack [ top ] - 1;
21         stack [ ++ top ] = i;
22     }
23     //求右边界
24     top = 0;
25     stack [ 0 ] = n + 1;
26     h [n + 1 ] = - 1;
27     for ( int i = n; i >= 1; i --)
28     {
29         while ( h [ stack [ top ]] >= h [ i ])
30             top --;
31         r [ i ] = stack [ top ] - i - 1;
32         stack [ ++ top ] = i;
33     }
34     int ans = - 1;
35     for ( int i = 1; i <= n; i ++)
36         ans = max( ans , ( l [ i ] + r [ i ] + 1) * h [ i ]);
37        
38     return ans;
39 }
40
41 int main()
42 {
43     while ( scanf( "%d%d" , & m , &n) != EOF)
44     {
45         memset( g , 0 , sizeof( g));
46         for ( int i = 1; i <= m; i ++)
47             for ( int j = 1; j <= n; j ++)
48             {
49                 scanf( "%d" , & g [ i ][ j ]);
50                 if ( g [ i ][ j ] == 1)
51                     g [ i ][ j ] += g [ i - 1 ][ j ];
52             }
53         int ans = - 1;
54         for ( int i = 1; i <= m; i ++)
55             ans = max( ans , work( g [ i ]));
56         printf( "%d \n " , ans);
57     }
58 }