codeforces 598D 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

准备写并查集的题 然后找了http://acm.hust.edu.cn/vjudge/contest/126073 这个题集

发现第一道题没想到用并查集 就用dfs写了#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define maxn 1005
using namespace std;
int dx[4]={0,0,-1,1},dy[4]={1,-1,0,0};
int n,m,k;
char maze[maxn][maxn];
int num[maxn][maxn];
int a[maxn*maxn];
int cnt;
int id=0;
void dfs(int x,int y)
{
num[x][y]=id;
for(int i=0;i<4;i++)
{
int nx=x+dx[i],ny=y+dy[i];
if(nx>=0&&nx<n&&ny>=0&&ny<m&&!num[nx][ny])
{
if(maze[nx][ny]=='*')
{
cnt++;}
else
{
dfs(nx,ny);
}
}
}
}
int main()
{
freopen("E:\\in.txt","r",stdin);
scanf("%d%d%d",&n,&m,&k);
memset(num,0,sizeof(num));
memset(a,0,sizeof(a));
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
cin>>maze[i][j];
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(maze[i][j]=='.'&&!num[i][j])//又标记又记录 学习了大牛的写法
{
id++;
cnt=0;
dfs(i,j);
a[id]=cnt;
}
}
while(k--)
{
int x,y;
cin>>x>>y;
cout<<a[num[x-1][y-1]]<<endl;
}
return 0;
}