hdu2870_Largest Submatrix

Posted on 2010-12-11 20:35 小小菜鸟蛋 阅读(450) 评论(0)  编辑 收藏 引用 所属分类: 某蛋的解题报告
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 }

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