zoukankan      html  css  js  c++  java
  • HDU 2870 Largest Submatrix(DP)

    题意  求最大相同字符子矩阵  其中一些字符可以转换

    其实就是HDU1505 1506的加强版  但是分了a,b,c三种情况  看哪次得到的面积最大

    对于某一个情况  可以把该字符和可以转换为该字符的位置赋值0 其它位置赋值1 这样就转化成了求最大全0矩阵的问题了

    对于转换后矩阵中的每个点 看他向上有多少个连续0 把这个值存在h数组中 再用l数组和r数组记录h连续大于等于该位置的最左边位置和最右位置 这样包含(i,j)点的最大矩阵面积就是(r[i][j]-l[i][j]+1)*h[i][j] 面积最大的点就是最大全0矩阵了

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int N = 1005;
    char s[N][N];
    int mat[N][N], l[N][N], r[N][N], h[N][N], n, m, ans;
    
    void makeMat (char a, char b, char c)
    {
        memset (mat, 0, sizeof (mat));
        memset (l, 0, sizeof (l));
        memset (r, 0, sizeof (r));
        memset (h, 0, sizeof (h));
        for (int i = 1; i <= n; ++i)
        {
            h[i][0] = h[i][m + 1] = -1;
            for (int j = 1; j <= m; ++j)
            {
                l[i][j] = r[i][j] = j;
                if (s[i][j] == a || s[i][j] == b || s[i][j] == c)  mat[i][j] = 1;
                if (mat[i][j] == 0) h[i][j] = h[i - 1][j] + 1;
            }
        }
    }
    
    int solve (char a, char b, char c)
    {
        makeMat (a, b, c);
        int aans = 0;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = m; j >= 1; --j)
                while (h[i][r[i][j] + 1] >= h[i][j])
                    r[i][j] = r[i][r[i][j] + 1];
    
            for (int j = 1; j <= m; ++j)
            {
                while (h[i][l[i][j] - 1] >= h[i][j])
                    l[i][j] = l[i][l[i][j] - 1];
                aans = max (aans, (r[i][j] - l[i][j] + 1) * h[i][j]);
            }
        }
        return aans;
    }
    
    int main()
    {
        while (~scanf ("%d%d", &n, &m))
        {
            for (int i = 1; i <= n; ++i)
                scanf ("%s", s[i] + 1);
            ans = max (solve ('x', 'b', 'c'), solve ('a', 'y', 'c'));
            ans = max (ans, solve ('a', 'b', 'w'));
            printf ("%d
    ", ans);
        }
        return 0;
    }

    Largest Submatrix

    Problem Description
    Now here is a matrix with letter 'a','b','c','w','x','y','z' and you can change 'w' to 'a' or 'b', change 'x' to 'b' or 'c', change 'y' to 'a' or 'c', and change 'z' to 'a', 'b' or 'c'. After you changed it, what's the largest submatrix with the same letters you can make?
     

    Input
    The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 1000) on line. Then come the elements of a matrix in row-major order on m lines each with n letters. The input ends once EOF is met.
     

    Output
    For each test case, output one line containing the number of elements of the largest submatrix of all same letters.
     

    Sample Input
    2 4 abcw wxyz
     

    Sample Output
    3
     


  • 相关阅读:
    各大型邮箱smtp服务器及端口收集
    阿里云邮箱POP3、SMTP设置教程
    thinkphp使用PHPMailer发送邮件
    SMTP错误码建议解决方法
    又来杭州了
    外企
    陈志武竞然是date的独立董事
    今天在当当上买了几本书,内有51反利的优惠券,也是一个suprise,
    工作 心态
    中国互联网创业,最好的城市是哪里?
  • 原文地址:https://www.cnblogs.com/acvay/p/3947289.html
Copyright © 2011-2022 走看看