codeforces 919C Seat Arrangements 思维模拟

C. Seat Arrangements

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Suppose that you are in a campus and have to go for classes day by day. As you may see, when you hurry to a classroom, you surprisingly find that many seats there are already occupied. Today you and your friends went for class, and found out that some of the seats were occupied.

The classroom contains n rows of seats and there are m seats in each row. Then the classroom can be represented as an n × m matrix. The character '.' represents an empty seat, while '*' means that the seat is occupied. You need to find k consecutive empty seats in the same row or column and arrange those seats for you and your friends. Your task is to find the number of ways to arrange the seats. Two ways are considered different if sets of places that students occupy differs.

Input

The first line contains three positive integers n, m, k (1 ≤ n, m, k ≤ 2 000), where n, m represent the sizes of the classroom and k is the number of consecutive seats you need to find.

Each of the next n lines contains m characters '.' or '*'. They form a matrix representing the classroom, '.' denotes an empty seat, and '*' denotes an occupied seat.

Output

A single number, denoting the number of ways to find k empty seats in the same row or column.

Examples

input

2 3 2
**.
...

output

3

input

1 2 2
..

output

1

input

3 3 4
.*.
*.*
.*.

output

0

Note

In the first sample, there are three ways to arrange those seats. You can take the following seats for your arrangement.

• (1, 3), (2, 3)
• (2, 2), (2, 3)
• (2, 1), (2, 2)

k个人必须连续坐，'.'代表空位，有多少种方法让k个人连续坐下

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <set>
#define debug(a)  cout << #a << " = "  << a <<endl
using namespace std;
int main() {
    int n,m,k,vis[];
    while( cin >> n >> m >> k ) {
        memset(vis,,sizeof(vis));
        int ans = ;
        for(int i=;i<n;i++) {
            int tmp = ;
            for(int j=;j<m;j++) {
                char c;
                cin >> c;
                if( c == '.' ) {
                    vis[j] ++; //记录每一列连续空位的个数
                    tmp ++;    //记录每一行连续空位的个数
                } else {       //当此处不为空位时，关于这个空位的行的连续、列的连续均被破坏
                    vis[j] = ;
                    tmp = ;
                }
                if( tmp >= k ) {
                    ans ++;
                }
                if( vis[j] >= k && k !=  ) { //当k为1的时候行的连续与列的连续会重复一次，所以特判k=1
                    ans ++;
                }
            }
        }
        cout << ans << endl;
    }
    return ;
}