Codeforces Round #460 (Div. 2)C. Seat Arrangements

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在一次校园课堂中，你需要为你和朋友们找到一串连续的空座位。教室为n行m列，每个座位用'.'表示空位，'*'表示已占位。目标是找到k个连续的空位，可以是同一行或同一列。给定教室布局，计算满足条件的座位排列方式的数量。题目提供n、m和k的值，以及教室的座位状态。第一样本有三种不同的安排方式。

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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)

参考了师兄的代码。。。

题目大意是一间课室里的座位 . 是空位 * 是已被占座，有k个同学要挨着坐，横竖都可以。一共有多少种坐的方法。

因为输入是没有空格的，所以要先用string再转为二维数组。还有k==1时答案要除以2。

#include <iostream>
#include<string.h>
using namespace std;
int main(){
	int n,m,k;cin>>n>>m>>k;
	string s;
	char a[n][m]={0};
	int ans=0;
	for(int i=0;i<n;i++){
		cin>>s;
		for(int j=0;j<m;j++){
			a[i][j]=s[j];
		}
	}
if(m==1&&n==1&&k==1){
	if(s=="*")cout<<0;
	else cout<<1;
	return 0;
}

	for(int i=0;i<n;i++){
	
	int com=0;
		for(int j=0;j<m;j++){
			   
			   if(a[i][j]=='.'){
					com++;
					if(com>=k)
					ans++;
				}
				else
				com=0;
			
		}
	}
	
		for(int i=0;i<m;i++){
		int com=0;
		for(int j=0;j<n;j++){
			
		
				if(a[j][i]=='.'){
					com++;
					if(com>=k)
					ans++;
				}
				else
				com=0;
			
		}
	}
	
	if(k==1)ans/=2;
	cout<<ans;
	
	return 0;
}