POJ 1979：Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 26058		Accepted: 14139

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile

'#' - a red tile

'@' - a man on a black tile(appears exactly once in a data set)

The end of the input is indicated by a line consisting of two zeros.

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

...@...

###.###

..#.#..

..#.#..

0 0

Sample Output

给一张图，@为起始点，'.'能走，‘#’不能走，问一共能走到多少'.'。

在深夜能做到这种水题也真是很令人高兴的事情。

深度搜索水题。

代码：

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

#include <cstring>

#pragma warning(disable:4996)

using namespace std;

int row,col,sum;

char value[30][30];

int flag[30][30];

void dfs(int x,int y)

{

	flag[x][y]=1;

	if(x>1&&flag[x-1][y]==0&&value[x-1][y]=='.')

	{

		dfs(x-1,y);

	}

	if(y>1&&flag[x][y-1]==0&&value[x][y-1]=='.')

	{

		dfs(x,y-1);

	}

	if(x<row&&flag[x+1][y]==0&&value[x+1][y]=='.')

	{

		dfs(x+1,y);

	}

	if(y<col&&flag[x][y+1]==0&&value[x][y+1]=='.')

	{

		dfs(x,y+1);

	}

}

void solve1()

{

	int i,j;

	for(i=1;i<=row;i++)

	{

		for(j=1;j<=col;j++)

		{

			if(value[i][j]=='@')

			{

				dfs(i,j);

				return;

			}

		}

	}

}

void solve2()

{

	int i,j;

	for(i=1;i<=row;i++)

	{

		for(j=1;j<=col;j++)

		{

			if(flag[i][j])

			{

				sum++;

			}

		}

	}

}

int main()

{

	int i,j;

	while(cin>>col>>row)

	{

		if(col+row==0)

			break;

		memset(flag,0,sizeof(flag));

		sum=0;

		for(i=1;i<=row;i++)

		{

			cin>>value[i]+1;

		}

		solve1();

		solve2();

		cout<<sum<<endl;

	}

	return 0;

}

巴特西

POJ 1979：Red and Black

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