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原题链接：

一：原题内容

2 4abcwwxyz

3

二：分析理解

这题用栈的方法竟然超时了，可能是stack的内存分配时间大了，只能改成比较麻烦的定义两个数组，left和right。。。。。。

这题是hdu1506-->

和hdu1505--> 的进阶版

三：AC代码

#include
   
      #include
    
     #include
     
        using namespace std;char ch[1005][1005];int a[1005][1005];int b[1005][1005];int c[1005][1005];int l[1005];int r[1005];int n, m;int Solve(int x[1005][1005]){	int finalAns = -1;	for (int i = 1; i <= n; i++)	{		l[1] = 1;		r[m] = m;		for (int j = 2; j <= m; j++)		{			l[j] = j;			while (l[j] != 1 && x[i][j] <= x[i][l[j] - 1])				l[j] = l[l[j] - 1];		}		for (int j = m - 1; j >= 1; j--)		{			r[j] = j;			while (r[j] != m&&x[i][j] <= x[i][r[j] + 1])				r[j] = r[r[j] + 1];		}		for (int j = 1; j <= m; j++)			if (x[i][j] * (r[j] - l[j] + 1) > finalAns)				finalAns = x[i][j] * (r[j] - l[j] + 1);	}	return finalAns;}int main(){	char ch;	while (~scanf("%d%d", &n, &m))	{		for (int i = 1; i <= n; i++)		{			getchar();			for (int j = 1; j <= m; j++)			{				scanf("%c", &ch);				//a				if (ch == 'a' || ch == 'w' || ch == 'y' || ch == 'z')					a[i][j] = a[i - 1][j] + 1;				else					a[i][j] = 0;				//b				if (ch == 'b' || ch == 'w' || ch == 'x' || ch == 'z')					b[i][j] = b[i - 1][j] + 1;				else					b[i][j] = 0;				//c				if (ch == 'c' || ch == 'x' || ch == 'y' || ch == 'z')					c[i][j] = c[i - 1][j] + 1;				else					c[i][j] = 0;			}		}		int a_ans = Solve(a);		int b_ans = Solve(b);		int c_ans = Solve(c);		printf("%d\n", max(max(a_ans, b_ans), c_ans));	}	return 0;}
     
    
   

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