hdu 2870 Largest Submatrix

最新推荐文章于 2021-07-30 19:13:48 发布

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#动态规划-子矩阵

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本文介绍了一种算法，用于解决给定可替换字符的矩阵中寻找最大相同字母子矩阵的问题。通过将可替换字符统一替换并计算每种情况下的最大子矩阵面积，最终输出最大的子矩阵元素数量。

 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

这道题直接把可替换的分别全换为a,b,c然后计算分别由a,b,c最大子矩阵的面积,此处求法可参照http://blog.youkuaiyun.com/ffgcc/article/details/78491268,这里我们多次对每一行求解即可.

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e3+10;
char app[N][N];
int arr[N][N];
int dp[N][N];
int num[N];
int l[N];
int r[N];
int n,m;
int fi()
{
    memset(dp,0,sizeof dp);
    int mx=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(arr[i][j]==1)
                dp[i][j]=dp[i-1][j]+1;
            else
                dp[i][j]=0;
            num[j]=dp[i][j];
        }
        l[0]=0;r[m-1]=m-1;

        for(int j=1;j<m;j++)
        {
            int tt=j;
            while(tt>0&&num[j]<=num[tt-1]) tt=l[tt-1];
            l[j]=tt;
        }
        for(int j=m-1;j>=0;j--)
        {
            int tt=j;
            while(tt<m-1&&num[j]<=num[tt+1])tt=r[tt+1];
            r[j]=tt;
        }
        for(int j=0;j<m;j++)
            mx=max((r[j]-l[j]+1)*num[j],mx);
    }
    return mx;

}
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>m&&n)
    {
        for(int i=1;i<=n;i++)
            cin>>app[i];
        memset(arr,0,sizeof arr);
        int mx=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<m;j++)
                if(app[i][j]=='w'||app[i][j]=='y'||app[i][j]=='z'||app[i][j]=='a')
                    arr[i][j]=1;
        }
        //cout<<fi()<<endl;
        mx=max(mx,fi());
        memset(arr,0,sizeof arr);
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<m;j++)
                if(app[i][j]=='w'||app[i][j]=='x'||app[i][j]=='z'||app[i][j]=='b')
                    arr[i][j]=1;
        }
        mx=max(mx,fi());
                //cout<<fi()<<endl;
        memset(arr,0,sizeof arr);
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<m;j++)
                if(app[i][j]=='x'||app[i][j]=='y'||app[i][j]=='z'||app[i][j]=='c')
                arr[i][j]=1;
        }
                //cout<<fi()<<endl;
        mx=max(mx,fi());
        cout<<mx<<endl;
    }
    return 0;
}