codeforces_598D. Igor In the Museum(dfs)

D. Igor In the Museum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Igor is in the museum and he wants to see as many pictures as possible.

Museum can be represented as a rectangular field of n × m cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are marked with '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.

At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.

For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picture he can see.

Input

First line of the input contains three integers n, m and k (3 ≤ n, m ≤ 1000, 1 ≤ k ≤ min(n·m, 100 000)) — the museum dimensions and the number of starting positions to process.

Each of the next n lines contains m symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.

Each of the last k lines contains two integers x and y (1 ≤ x ≤ n, 1 ≤ y ≤ m) — the row and the column of one of Igor's starting positions respectively. Rows are numbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells.

Output

Print k integers — the maximum number of pictures, that Igor can see if he starts in corresponding position.

Examples

input

5 6 3
******
*..*.*
******
*....*
******
2 2
2 5
4 3

output

6
4
10

input

4 4 1
****
*..*
*.**
****
3 2

output

8

每个连通块中从任意一个位置开始能看到的图片数是一样的，所以先求出每个连通块能看到的图片的个数，同时每个'.'指向其所对应的连通块，最后查询一下就能得到答案。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
char a[1010][1010];
int  v1[1010][1010];
int  v2[1010][1010][4];
int c[1010][1010];
int sum=0,top=0;
int bank[1000010];

int dfs(int x,int y)
{
    v1[x][y]=1;
    c[x][y]=top;
    for(int i=0;i<4;i++)
    {
        int xx=x+dx[i];
        int yy=y+dy[i];
        if(a[xx][yy]=='*'&&v2[xx][yy][i]==0)
        {
            v2[xx][yy][i]=1;
            sum++;continue;
        }
        else if(a[xx][yy]=='*')
        {
            continue;
        }
        else if(a[xx][yy]=='.'&&v1[xx][yy]==0)
        {
            dfs(xx,yy);
        }
        else
        {
            continue;
        }
    }
}

int main()
{
    int i,j,n,m,k;
    Si(n);Si(m);Si(k);
    for(i=1;i<=n;i++)
    {
        Ss(a[i]+1);
    }
    mem(v1,0);
    mem(v2,0);
    mem(c,0);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(a[i][j]=='.'&&v1[i][j]==0)
            {
                sum=0;
                dfs(i,j);
                bank[top]=sum;
                top++;
            }
        }
    }
    W(k)
    {
        int x,y;
        Si(x);Si(y);
        Pi(bank[c[x][y]]);
    }

    return 0;
}