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

C. Seat Arrangements

time limit per test1 second

memory limit per test256 megabytes

Problem Description

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)

解题心得：

题意就是说，在同一行或者同一列中要找k个连续的空地，有多少种找法。
思路是先找每一行，将连续的空地加起来，如果大于等于k个，就用连续空地数-k+1，先将所有的行找出来，再将所有的列找出来。就这个思路，被hack了很多次，就是处理k==1的问题啊。第一次被hack，想将列中重复找的k==1给除去，然后又被hack，没除完，最后k==1直接特判一个一个的数空地就过了。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 2010;

char maps[maxn][maxn];

int n,m,k;

long long ans = 0;

void init()

{

    for(int i=0;i<n;i++)

        scanf("%s",maps[i]);

}

void check_row()//找每一行中的连续空地数目

{

    for(int i=0;i<n;i++)

        for(int j=0;j<m;j++)

            if(maps[i][j] == '.')

            {

                int z = j+1;

                int tot = 1;

                while(maps[i][z] == '.')

                {

                    tot++;

                    z++;

                }

                if(tot >= k)

                    ans += tot-k+1;

                j = z-1;

            }

}

void check_col()//找每一列中的连续的空地数目

{

    for(int i=0;i<m;i++)

        for(int j=0;j<n;j++)

            if(maps[j][i] == '.')

            {

                int z = j+1;

                int tot = 1;

                while(maps[z][i] == '.')

                {

                    z++;

                    tot++;

                }

                if(tot == 1)

                    continue;

                if(tot >= k)

                    ans += tot-k+1;

                j = z-1;

            }

}

void special_judge(){//特判k==1的时候的方案数

    int ans = 0;

    for(int i=0;i<n;i++)

        for(int j=0;j<m;j++)

            if(maps[i][j] == '.')

                ans++;

    printf("%d\n",ans);

}

int main()

{

    scanf("%d%d%d",&n,&m,&k);

    init();

    if(k == 1){

        special_judge();

        return 0;

    }

    check_row();

    check_col();

    printf("%lld\n",ans);

    return 0;

}

巴特西