HDU 1312:Red and Black（DFS搜索）

 

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 ;
}