2016SDAU编程练习二1016

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Red and Black

Problem 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) 

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
45 59 6 13 

Source
Asia 2004, Ehime (Japan), Japan Domestic

题意：从@走四个方向，碰到#和边界返回

思路：深搜广搜应该都行吧

感想：这算比较简单的了

AC代码：

#include <iostream>
#include<string.h>
using namespace std;
#define N 20
int m,n;

int dirctions[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int visited[N][N];
char map[N][N];
int sum=0;
void DFS(int x,int y)
{

int nx,ny;
for(int i=0;i<4;i++)
{
nx=x+dirctions[i][0];
ny=y+dirctions[i][1];

if(map[nx][ny]=='.'&&nx>=0&&nx<m&&ny>=0&&ny<n&&visited[nx][ny]==0)
{

visited[nx][ny]=1;
sum++;
DFS(nx,ny);

}

}

}


int main()
{
int i,j,x,y;

while(cin>>n>>m)
{
if(m==0||n==0)
break;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
{
cin>>map[i][j];
if(map[i][j]=='@')
{
x=i;
y=j;
}
}

memset(visited,0,sizeof(visited));
sum=1;
DFS(x,y);
cout<<sum<<endl;

}

return 0;
}

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