标签：dp   hdu   动态规划   最大全0矩阵   

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\n", ans);
    }
    return 0;
}

Largest Submatrix

2 4
abcw
wxyz

3

HDU 2870 Largest Submatrix(DP),布布扣,bubuko.com

HDU 2870 Largest Submatrix(DP)

标签：dp   hdu   动态规划   最大全0矩阵   

原文地址：http://blog.csdn.net/iooden/article/details/38491987