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

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)

思路：

最暴力的是判断每个位置横竖是否可行，然而会超时。

又想了，记录横竖有多少个连续的空座位，使其 -k+1，k==1也特判了，还是有未知的bug，遂放弃重写。

最近在刷DP基础，用着DP的思想写了出来，其实与上面那种方法相差无几。

分别在横竖上进行dp，空座位用1标记，否则用0。

如果两个都是空座位，那么前一个变0，这个变为前一个+1。最后遍历求解，（特判k==1）。

详见代码

代码：

#include<stdio.h>

int col[5005][5005],mul[5005][5005];

int main()
{
int n,m,k,i,j,ans=0;
char s[2005];
scanf("%d%d%d",&n,&m,&k);
for(i=0;i<n;i++)
{
scanf("%s",s);
for(j=0;j<m;j++)
if(s[j]=='.')col[i][j]=mul[i][j]=1;
}

for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(mul[i][j]&&mul[i+1][j])mul[i+1][j]+=mul[i][j],mul[i][j]=0;

if(col[i][j]&&col[i][j+1])col[i][j+1]+=col[i][j],col[i][j]=0;
}

for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(col[i][j]-k+1>0)ans+=col[i][j]-k+1;
if(mul[i][j]-k+1>0)ans+=mul[i][j]-k+1;
}
}

if(k==1)ans/=2;
printf("%d\n",ans);

return 0;
}