hdu2870 Largest Submatrix 1506 1505

最新推荐文章于 2019-04-16 20:39:31 发布

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最新推荐文章于 2019-04-16 20:39:31 发布

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本文探讨了在矩阵中通过变换特定字符以形成最大相同字母子矩阵的问题，并提供了求解方法。

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Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1411 Accepted Submission(s): 670

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

将全部字符依次转化a, b, c, 再分别求出这三个矩阵的最大子矩阵即可.

思路：枚举后转换为 hdu 1505 hdu 1506 的情况。

可以转化为a的有：w，y， z

可以转化为b的有：w，x， z

可以转换为c的有：x， y， z

所以

a, w, y, z一组

b, w, x, z一组

c, x, y, z一组

分别dp

///对于每一个a[i],用dp找出a[i]左边和右边连续大于自己的数的长度    
///l[i]表示比a[i]大的数连续的最左边的位置    
///r[i]表示比a[i]大的数连续的最右边的位置  
#include<iostream>    
#include<cstring>    
#include<algorithm>  
using namespace std;    
int l[1005],r[1005];   
int MAX;
char str[1005][1005];  
int h[1005][1005];
int m,n,MAXANS;
void DP()
{
	int i;
	for(int x=1;x<=m;++x)
	{
		l[1]=1;
		r[n]=n;
		for(i=2;i<=n;++i)  
		{  
			int t=i;  
			while(t>1&&h[x][i]<=h[x][t-1])  
				t=l[t-1];  
			l[i]=t;  
		} 

		for(i=n-1;i>=1;--i)  
		{  
			int t=i;  
			while(t<n&&h[x][i]<=h[x][t+1])  
				t=r[t+1];  
			r[i]=t;  
		} 

		MAX=0; 
		for(i=1;i<=n;++i)  
			if((r[i]-l[i]+1)*h[x][i]>MAX)  
				MAX=(r[i]-l[i]+1)*h[x][i]; 
		if(MAX>MAXANS)
			MAXANS=MAX;
	}  
}
int main()    
{  
	while(cin>>m>>n)  
	{  
		int i,j;  
		for(i=1;i<=m;++i)  
			cin>>str[i]+1;  
		MAXANS=0;
		for(i=1;i<=m;++i)
		{
			for(j=1;j<=n;++j)
			{
				if(str[i][j]=='a'||str[i][j]=='w'||str[i][j]=='z'||str[i][j]=='y')
					h[i][j]=h[i-1][j]+1;
				else
					h[i][j]=0;
			}
		}
		DP();
		for(i=1;i<=m;++i)
		{
			for(j=1;j<=n;++j)
			{
				if(str[i][j]=='b'||str[i][j]=='w'||str[i][j]=='z'||str[i][j]=='x')
					h[i][j]=h[i-1][j]+1;
				else
					h[i][j]=0;
			}
		}
		DP();
		for(i=1;i<=m;++i)
		{
			for(j=1;j<=n;++j)
			{
				if(str[i][j]=='c'||str[i][j]=='x'||str[i][j]=='y'||str[i][j]=='z')
					h[i][j]=h[i-1][j]+1;
				else
					h[i][j]=0;
			}
		}
		DP();
		cout<<MAXANS<<endl;  
	}  
}