codeforces 598D - Igor In the Museum

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

这是一般的搜索题，当你发现超时的时候，你就少了一个优化，可以把连接起来的'.',标记为一个区域，这样下次再碰到这个区域的点时，可以直接输出，避免再次

搜索而浪费时间，注意num[]数组开的大小，我前面是num[1100],后来才明白，如果n=1000,m=1000; 那么区域的个数就不是1100那么少了，这个要注意，我就

是后来才发现的。。。。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

char map[1100][1100];
int vis[1100][1100];
int sum;
int time[1100][1100];
int num[1100*1100];
int n,m;
int b[4][2]={ {0,1},{0,-1},{1,0},{-1,0} };

int dfs(int x,int y,int pos)
{
int i;
vis[x][y]=1;
for(i=0;i<4;i++)
{
int dx=x+b[i][0];
int dy=y+b[i][1];
if(vis[dx][dy]==0&&dx>=1&&dx<=n&&dy>=1&&dy<=m)
{
if(map[dx][dy]=='*')
{
sum++;
}
else
{
time[dx][dy]=pos;
dfs(dx,dy,pos);
}
}
}
return 0;
}

int main()
{
int i,j;
int k;
int pos;
scanf("%d%d%d",&n,&m,&k);
for(i=1;i<=n;i++)
{
getchar();
for(j=1;j<=m;j++)
{
scanf("%c",&map[i][j]);
vis[i][j]=0;
}
}
getchar();
pos=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(map[i][j]=='.'&&vis[i][j]==0)
{
pos++;
sum=0;
time[i][j]=pos;
dfs(i,j,pos);
num[time[i][j]]=sum;
}
}
}
while(k--)
{
int sx,sy;
scanf("%d%d",&sx,&sy);
printf("%d\n",num[time[sx][sy]]);
}
return 0;
}