HDOJ2870 Largest Submatrix

标签：ges   复制   name   垂直   初始   main   getchar   大小   for   

一道\(DP\)

原题链接

发现只有\(a,b,c\)三种情况，所以直接初始化成三个\(01\)方阵，找最大子矩阵即可。
我是先初始化垂直上的高度，然后对每一行处理出每个点向左向右的最大延伸，并不断计算矩阵大小来更新答案。
因为不想开函数传数组，所以全写在主函数复制粘贴了三遍。。代码显得比较冗长。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1010;
int A[N][N], B[N][N], C[N][N], l[N], r[N];
char re_l()
{
    char c = getchar();
    for (; c != 'a'&&c != 'b'&&c != 'c'&&c != 'x'&&c != 'y'&&c != 'z'&&c != 'w'; c = getchar());
    return c;
}
inline int maxn(int x, int y)
{
    return x > y ? x : y;
}
int main()
{
    int i, j, ma, n, m;
    char c;
    while (scanf("%d%d", &n, &m)==2)
    {
        ma = 1;
        memset(A, 0, sizeof(A));
        memset(B, 0, sizeof(B));
        memset(C, 0, sizeof(C));
        for (i = 1; i <= n; i++)
            for (j = 1; j <= m; j++)
            {
                c = re_l();
                if (c == 'a' || c == 'w' || c == 'y' || c == 'z')
                    A[i][j] = A[i - 1][j] + 1;
                if (c == 'b' || c == 'w' || c == 'x' || c == 'z')
                    B[i][j] = B[i - 1][j] + 1;
                if (c == 'c' || c == 'x' || c == 'y' || c == 'z')
                    C[i][j] = C[i - 1][j] + 1;
            }
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= m; j++)
                l[j] = r[j] = j;
            A[i][0] = A[i][m + 1] = -1;
            for (j = 1; j <= m; j++)
                while (A[i][j] <= A[i][l[j] - 1])
                    l[j] = l[l[j] - 1];
            for (j = m; j; j--)
                while (A[i][j] <= A[i][r[j] + 1])
                    r[j] = r[r[j] + 1];
            for (j = 1; j <= m; j++)
                ma = maxn(ma, (r[j] - l[j] + 1)*A[i][j]);
        }
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= m; j++)
                l[j] = r[j] = j;
            B[i][0] = B[i][m + 1] = -1;
            for (j = 1; j <= m; j++)
                while (B[i][j] <= B[i][l[j] - 1])
                    l[j] = l[l[j] - 1];
            for (j = m; j; j--)
                while (B[i][j] <= B[i][r[j] + 1])
                    r[j] = r[r[j] + 1];
            for (j = 1; j <= m; j++)
                ma = maxn(ma, (r[j] - l[j] + 1)*B[i][j]);
        }
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= m; j++)
                l[j] = r[j] = j;
            C[i][0] = C[i][m + 1] = -1;
            for (j = 1; j <= m; j++)
                while (C[i][j] <= C[i][l[j] - 1])
                    l[j] = l[l[j] - 1];
            for (j = m; j; j--)
                while (C[i][j] <= C[i][r[j] + 1])
                    r[j] = r[r[j] + 1];
            for (j = 1; j <= m; j++)
                ma = maxn(ma, (r[j] - l[j] + 1)*C[i][j]);
        }
        printf("%d\n", ma);
    }
    return 0;
}

ps：因为我用的是\(VS\)，所以代码会自动补空格，而我本来的风格是没有空格的。

HDOJ2870 Largest Submatrix

标签：ges   复制   name   垂直   初始   main   getchar   大小   for   

原文地址：https://www.cnblogs.com/Iowa-Battleship/p/9519810.html