HDOJ1312 Red and black(DFS深度优先搜索)

朴灿烈づ我的快乐病毒、 2021-12-17 14:07 210阅读 0赞

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

#include<stdio.h>
    #include<string.h>
    using namespace std;
    char s[50][50];
    int n,m;
    int dfs(int x,int y)
    {
        int i,j,k;
        if(x<0||x>n-1||y<0||y>m-1)
            return 0;
        
        if(s[x][y]=='#')
        {
            return 0;
        }    
        else
        {
            s[x][y]='#';
            return 1+dfs(x+1,y)+dfs(x-1,y)+dfs(x,y+1)+dfs(x,y-1);//搜索四个方向
        }
    }
    int main() 
    {
        int i,j,k;
        while(scanf("%d%d",&m,&n)!=EOF)
        {
            if(m==0&&n==0)
                break;
            for(i=0;i<n;i++)
            {
                scanf("%s",s[i]);
            }
            int sum=0;
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    if(s[i][j]=='@')
                    {
                        sum=dfs(i,j);    
                    } 
                } 
            }
            printf("%d\n",sum);
        }
        return 0;
    }

永不言弃！和ACM死磕到底！

转载于:https://www.cnblogs.com/jianqiao123/p/10780499.html