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.

Sample test(s)

input

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

output

6
4
10

input

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

output

8

题意就是.和*相交的那部分有画，问每一个.连通块能看到多少幅画。

简单深搜，这题的一个亮点在于归类，将一个连通块的点用一个id标记一下，mark一下。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;

int n, m, k, id;
int classify[1000005];
int result[1005][1005];
int vis[1005][1005];
char val[1005][1005];

int dfs(int x, int y)
{
	vis[x][y] = id;//标记一个连通块内的点
	if (val[x + 1][y] == '*')
	{
		result[x][y]++;
	}
	if (val[x - 1][y] == '*')
	{
		result[x][y]++;
	}
	if (val[x][y + 1] == '*')
	{
		result[x][y]++;
	}
	if (val[x][y - 1] == '*')
	{
		result[x][y]++;
	}
	if (vis[x - 1][y] == 0 && val[x - 1][y] == '.')
	{
		result[x][y] = dfs(x - 1, y) + result[x][y];
	}
	if (vis[x + 1][y] == 0 && val[x + 1][y] == '.')
	{
		result[x][y] = dfs(x + 1, y) + result[x][y];
	}
	if (vis[x][y - 1] == 0 && val[x][y - 1] == '.')
	{
		result[x][y] = dfs(x, y - 1) + result[x][y];
	}
	if (vis[x][y + 1] == 0 && val[x][y + 1] == '.')
	{
		result[x][y] = dfs(x, y + 1) + result[x][y];
	}
	return result[x][y];
}

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);

	int i, j;
	int temp1, temp2;
	scanf("%d%d%d", &n, &m, &k);

	for (i = 1; i <= n; i++)
		cin >> val[i] + 1;
	id = 1;
	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= m; j++)
		{
			if (val[i][j] == '.'&&vis[i][j] == 0)
			{
				classify[id]= dfs(i, j);
				id++;
			}
		}
	}
	for (i = 1; i <= k; i++)
	{
		scanf("%d%d", &temp1, &temp2);
		printf("%d\n", classify[vis[temp1][temp2]]);
	}
	//system("pause");
	return 0;
}