hdu2870_Largest Submatrix

http://acm.hdu.edu.cn/showproblem.php?pid=2870

这题跟pku2796、2559、3494一样。。。

只可以转为a、b、c三种字母，所以直接枚举，枚举可以转为a的矩阵、b的矩阵、c的矩阵，然后得到每个矩阵的最大矩阵值，再取其中的最大值~~

01 #include <cstdio>
02 #include <cstring>
03 #include <algorithm>
04 using namespace std;
05
06 const int N = 1010;
07 int a [N ][N ], b [N ][N ], c [N ][N ];
08 int n , m;
09 int stack [N ], l [N ], r [N ];
10
11 int max( int x , int y , int z , int u)
12 {
13     return max( max( x , y ), max( z , u));
14 }
15
16 int work( int h [])
17 {
18     int top = 0;
19     stack [ 0 ] = 0;
20     h [ 0 ] = - 1;
21     for ( int i = 1; i <= n; i ++)
22     {
23         while ( h [ stack [ top ]] >= h [ i ])
24             top --;
25         l [ i ] = i - stack [ top ] - 1;
26         stack [ ++ top ] = i;
27     }
28     top = 0;
29     stack [ 0 ] = n + 1;
30     h [n + 1 ] = - 1;
31     for ( int i = n; i >= 1; i --)
32     {
33         while ( h [ stack [ top ]] >= h [ i ])
34             top --;
35         r [ i ] = stack [ top ] - i - 1;
36         stack [ ++ top ] = i;
37     }
38     int ans = - 1;
39     for ( int i = 1; i <= n; i ++)
40         ans = max( ans , ( r [ i ] + l [ i ] + 1) * h [ i ]);
41        
42     return ans;
43 }
44
45 int main()
46 {
47     while ( scanf( "%d%d" , & m , &n) != EOF)
48     {
49         memset( a , 0 , sizeof( a));
50         memset(b , 0 , sizeof(b));
51         memset( c , 0 , sizeof( c));
52         char ch;
53         for ( int i = 1; i <= m; i ++)
54         {
55             scanf( "%c" , & ch);  //读掉过行符
56             for ( int j = 1; j <= n; j ++)
57             {
58                 scanf( "%c" , & ch);
59                 if ( ch == 'w' || ch == 'y' || ch == 'z' || ch == 'a')
60                     a [ i ][ j ] = a [ i - 1 ][ j ] + 1;
61                 if ( ch == 'w' || ch == 'x' || ch == 'z' || ch == 'b')
62                     b [ i ][ j ] = b [ i - 1 ][ j ] + 1;
63                 if ( ch == 'x' || ch == 'y' || ch == 'z' || ch == 'c')
64                     c [ i ][ j ] = c [ i - 1 ][ j ] + 1;
65             }
66         }
67         int ans = - 1;
68         for ( int i = 1; i <= m; i ++)
69             ans = max( ans , work( a [ i ]), work(b [ i ]), work( c [ i ]));
70         printf( "%d \n " , ans);
71     }
72 }