Red and Black

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=67590#problem/D

 

Red and Black

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1312

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

 

Sample Output

45

59

6

13

 

CODE：

#include<stdio.h>

char a[25][25];

int w, h;

int f(int x, int y)
{
    if(x < 0 || x >= h || y < 0 || y >= w)
        return 0;
    if(a[x][y] == '#')
        return 0;
    else
    {
        a[x][y] = '#';
        return 1 + f(x, y-1) + f(x, y+1) + f(x-1, y) + f(x+1, y);
    }
}

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

    while(scanf("%d%d", &w, &h), w+h)
    {
        for(i = 0; i < h; i++)
            for(j = 0; j < w; j++)
            {
                scanf(" %c", &a[i][j]);
                if(a[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
            }

        printf("%d
", f(x, y));
    }
    return 0;
}

受不鸟为啥这样不对，先放这好了：

#include<iostream>
#include<cstdio>

using namespace std;

#define N 25

int w, h, cou;
char maps[N][N];
int dir[4][2] = { {-1, 0}, {0, -1}, {0, 1}, {1, 0}};

void DFS(int x, int y)
{
    int i, nx, ny;

    cou++;
    
    maps[x][y] = '#';

    for(i = 0; i < 4; i++)
    {
        nx = x + dir[i][0];
        ny = y + dir[i][1];
        if(maps[nx][ny] == '.' && nx >= 0 && nx < h && ny >= 0 && ny < w)
        {
            DFS(nx, ny);
        }
    }
}
int main()
{
    int i, j;

    while(cin >> w >> h, w+h)
    {
       for(i = 0; i < h; i++)
			for(j = 0; j < w; j++)
				scanf(" %c", &maps[i][j]);

        for(i = 0; i < h; i++)
            for(j = 0; j < w; j++)
        {
            if(maps[i][j] == '@')
                DFS(i, j);
        }
        cout << cou << endl;
    }
    return 0;
}