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

解题思路：

‘ 。 ’表示移动的位置，*表示墙，墙的每个面都用一幅画，我们要做的是，一个人在可移动的范围内最多能看到多少幅画，这个题不知道，在最后询问的过程中要用一维数组查找答案，其他的方法都是超时。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int m,n,k;
char map[1012][1012];
int xx,yy;
int id=0;
int vis[1012][1012];
int ans1[1012*1012];
int ans;

void DFS(int x,int y)
{
int i,j;
vis[x][y] = id;
for(j=0;j<4;j++)
{
if(!vis[x+dx[j]][y+dy[j]]&&x+dx[j]>=0&&x+dx[j]<n&&y+dy[j]>=0&&y+dy[j]<m)
{
if(map[x+dx[j]][y+dy[j]]=='*')
{
ans++;
}
else
DFS(x+dx[j],y+dy[j]);
}
}
}

int main()
{
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
id=0;
int i,j;
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
}
memset(vis, 0, sizeof(vis));
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(map[i][j]=='.'&&!vis[i][j])
ans=0,++id,DFS(i,j),ans1[id]=ans;
}
}

for(i=0;i<k;i++)
{
scanf("%d%d",&xx,&yy);
printf("%d\n",ans1[vis[xx-1][yy-1]]);
}

}
return 0;
}