Codeforces Round #662 (Div. 2) D. Rarity and New Dress
第一二层肯定是1，然后第三场在四个位置都符合要求就是2，然后这么一直往下，第（x,y）的值就是上面四个位置的值的最小值+1
(四个位置如下图）
（对于最下面的绿色的d来说，红色圈的就是它的四个位置
）
Carousel Boutique is busy again! Rarity has decided to visit the pony ball and she surely needs a new dress, because going out in the same dress several times is a sign of bad manners. First of all, she needs a dress pattern, which she is going to cut out from the rectangular piece of the multicolored fabric.
The piece of the multicolored fabric consists of $$$n \times m$$$ separate square scraps. Since Rarity likes dresses in style, a dress pattern must only include scraps sharing the same color. A dress pattern must be the square, and since Rarity is fond of rhombuses, the sides of a pattern must form a $$$45^{\circ}$$$ angle with sides of a piece of fabric (that way it will be resembling the traditional picture of a rhombus).
Examples of proper dress patterns: Examples of improper dress patterns: The first one consists of multi-colored scraps, the second one goes beyond the bounds of the piece of fabric, the third one is not a square with sides forming a $$$45^{\circ}$$$ angle with sides of the piece of fabric.
Rarity wonders how many ways to cut out a dress pattern that satisfies all the conditions that do exist. Please help her and satisfy her curiosity so she can continue working on her new masterpiece!
Input
The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 2000$$$). Each of the next $$$n$$$ lines contains $$$m$$$ characters: lowercase English letters, the $$$j$$$-th of which corresponds to scrap in the current line and in the $$$j$$$-th column. Scraps having the same letter share the same color, scraps having different letters have different colors.
Output
Print a single integer: the number of ways to cut out a dress pattern to satisfy all of Rarity's conditions.
Examples
Input
Copy
3 3 aaa aaa aaa
Output
Copy
10
Input
Copy
3 4 abab baba abab
Output
Copy
12
Input
Copy
5 5 zbacg baaac aaaaa eaaad weadd
Output
Copy
31
Note
In the first example, all the dress patterns of size $$$1$$$ and one of size $$$2$$$ are satisfactory.
In the second example, only the dress patterns of size $$$1$$$ are satisfactory.
#include
using namespace std;
typedef long long ll;
const ll maxn = 2050;
char a[maxn][maxn];ll dp[maxn][maxn];
check(ll x,ll y)
{
if(a[x][y]==a[x-1][y-1]&&a[x][y]==a[x-1][y]&&a[x][y]==a[x-1][y+1]&&a[x][y]==a[x-2][y])
{
return min(min(dp[x-1][y-1],dp[x-1][y]),min(dp[x-1][y+1],dp[x-2][y]))+1;
}
else return 1;
}
int main()
{
ll n,m;
scanf("%lld %lld",&n,&m);
for(ll i=1;i<=n;i++) scanf("%s",a[i]+1);
for(ll i=1;i<=m;i++) dp[1][i]=dp[2][i]=1;
for(ll i=3;i<=n;i++)
{
for(ll j=1;j<=m;j++)
{
dp[i][j]=check(i,j);
}
}
ll ans=0;
for(ll i=1;i<=n;i++)
{
for(ll j=1;j<=m;j++)
{
ans+=dp[i][j];
}
}
printf("%lld\n",ans);
}