E. Zbazi in Zeydabad

题目连接：

Description

A tourist wants to visit country Zeydabad for Zbazi (a local game in Zeydabad).

The country Zeydabad is a rectangular table consisting of n rows and m columns. Each cell on the country is either 'z' or '.'.

The tourist knows this country is named Zeydabad because there are lots of ''Z-pattern"s in the country. A ''Z-pattern" is a square which anti-diagonal is completely filled with 'z' and its upper and lower rows are also completely filled with 'z'. All other cells of a square can be arbitrary.

Note that a ''Z-pattern" can consist of only one cell (see the examples).

So he wants to count the number of ''Z-pattern"s in the country (a necessary skill for Zbazi).

Now your task is to help tourist with counting number of ''Z-pattern"s.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 3000) — the number of rows and columns respectively.

Each of the next n lines contains m characters 'z' or '.' — the description of Zeydabad.

Output

Print the only integer a — the number of ''Z-pattern"s in Zeydabad.

Sample Input

4 4

zzzz

zzz.

.z..

zzzz

Sample Output

16

Hint

题意

给你一个n*m的矩阵，然后问你里面有多少个z

题解：

数据结构

首先我们对于每一个点维护一个l[i][j],r[i][j]，表示这个点往左边最多延伸多少，往右边最多延伸多少

我们去维护每一个对角线就好了，我们从左下角跑到右上角

我们update(i,1)表示这个位置当前是可以的，他对于上面的i+r[i][j]-1个都是可以更新的

他这个点的贡献就是query(i)-query(i-len)，len = l[i][j]和i-last的最小值，显然他可以和下面len个进行组合

然后这样去跑就好了

我讲的比较抽象，还是看代码吧……

代码

#include
    
     using namespace std;const int maxn = 3e3+5;int a[maxn],n,m;char s[maxn][maxn];int l[maxn][maxn],r[maxn][maxn];vector
     
      over[maxn];int lowbit(int x){    return x&(-x);}void update(int x,int val){    x++;    for(int i=x;i
      
       =0;j--)            if(s[i][j]=='z')r[i][j]=r[i][j+1]+1;    }    long long ans = 0;    for(int i=0;i