HDU 2870 Largest Submatrix（最大完全子矩阵）

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本文介绍了一种算法，用于解决给定包含特定字符的矩阵中，通过替换某些字符后能够形成的最大的完全由相同字符组成的子矩阵的问题。该算法通过枚举目标字符并利用单调栈思想进行优化，实现了高效求解。

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2830 Accepted Submission(s): 1368

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

这个题就是分别枚举a，b，c；然后看谁的完全子矩阵最大，类似1506,1505；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int h[1005][1005];
char mat[1005][1005];
int r[1005];
int l[1005];
int n,m,ans;
void fun()
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            l[j]=r[j]=j;
        }
        h[i][0]=h[i][m+1]=-1;
        for(int j=1;j<=m;j++)
        {
            while(h[i][j]<=h[i][l[j]-1])
                l[j]=l[l[j]-1];
        }
        for(int j=m;j>=1;j--)
        {
            while(h[i][j]<=h[i][r[j]+1])
                r[j]=r[r[j]+1];
        }
        for(int j=1;j<=m;j++)
            ans=max(ans,h[i][j]*(r[j]-l[j]+1));
     } 
} 
int main()
{
    char ch;
    while(~scanf("%d%d",&n,&m))
    {
        ans=0;
        getchar();
        for(int i=1;i<=n;i++)
            scanf("%s",mat[i]+1);
        for(int i=1;i<=n;i++)
        {//枚举a 
            for(int j=1;j<=m;j++)
            {
                ch=mat[i][j];
                if(ch=='a'||ch=='w'||ch=='y'||ch=='z')
                    h[i][j]=h[i-1][j]+1;
                else h[i][j]=0;
            }
        }
        
        fun();
        for(int i=1;i<=n;i++)
        {//枚举b 
            for(int j=1;j<=m;j++)
            {
                ch=mat[i][j];
                if(ch=='b'||ch=='w'||ch=='x'||ch=='z')
                    h[i][j]=h[i-1][j]+1;
                else h[i][j]=0;
            }
        }
        fun();
        
        for(int i=1;i<=n;i++)
        {//枚举c 
            for(int j=1;j<=m;j++)
            {
                ch=mat[i][j];
                if(ch=='c'||ch=='x'||ch=='y'||ch=='z')
                    h[i][j]=h[i-1][j]+1;
                else h[i][j]=0;
            }
        }
        fun();
        printf("%d\n",ans);
    }
    return 0;
}