HDU 1312:Red and Black

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

 

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

45

59

6

13

题解：本题还是DFS搜索（上下左右），只是增加一个计数器，计算可以走的地方的个数。
      规定地图中有可通行的位置，也有不可通行的位置，已知起点，求起点的与它相连成一片的部分，在这道题里输出相连的位置的数目。
　   从起点开始，遍历每一个到达的点的四个方向（不再是八个），到达一个位置就将这个位置的字符变成不可走的'#'，并且计数+1。其实就是计数将可走变成不可走的操作进行了多少次。
这样可以不用担心走过了还会重复。
 
 AC
代码：如果你看过我的上一篇你一定会懂

#include<cstdio>

#include<cstring>

char pic[][];

int m,n,total;

int idx[][];

void dfs(int r,int c,int id)

{

    if(r<||r>=m||c<||c>=n)

        return;

    if(idx[r][c]==||pic[r][c]!='.')

        return;

    idx[r][c]=id;

    total++;

    for(int dr=-; dr<=; dr++)

        for(int dc=-; dc<=; dc++)

            if(dr==||dc==)

                dfs(r+dr,c+dc,id);

}

int main()

{

    int i,j;

    while(scanf("%d%d",&n,&m)==&&m&&n)

    {

        for(i =; i<m; i++)

            scanf("%s",pic[i]);

        memset(idx,,sizeof(idx));

        total=;

        for(i=; i<m; i++)

            for(j=; j<n; j++)

            {

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

                {

                    pic[i][j]='.';

                    dfs(i,j,);

                }

            }

        printf("%d\n",total);

    }

    return ;

}

 
 
 
这个是我在其他博客上看得到的方法，用#填充，可以一试！
 

#include <iostream>

using namespace std;

char a[][];

int n,m,total;

int dr[] = {,,,-};//行变化

int dc[] = {,,-,};//列变化

//上面的原来一直不会用，知道的话非常方便

bool judge(int x,int y)

{

    if(x< || x>n || y< || y>m)

        return ;

    if(a[x][y]=='#')

        return ;

    return ;

}

void dfs(int r,int c)

{

    total++;

    a[r][c]='#';               //走过一次，“。”变为“#”,避免重复

    for(int k=; k<; k++)

    {

        int lr = r + dr[k];

        int lc = c + dc[k];

        if(judge(lr,lc))

            continue;

        dfs(lr,lc);

    }

}

int main()

{

    while(cin>>m>>n&&m&&n)

    {

        int i,j,x,y;

        total=;

        for(i=; i<=n; i++)

            for(j=; j<=m; j++)

            {

                cin>>a[i][j];

                if(a[i][j]=='@')             //这里必须用变量x，y

                    x=i,y=j;

            }

        dfs(x,y);

        cout<<total<<endl;

    }

    return ;

}

 

HDU 1312:Red and Black（DFS搜索）的相关教程结束。