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  • Red and Black(DFS深搜实现)

    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
     
     
     
     
    深搜实现方案:
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<algorithm>
     4 #include<iostream>
     5 using namespace std;
     6 char map[110][110];
     7 int  vis[110][100];
     8 int  dir[8][2]={{0,1},{0,-1},{1,0},{-1,0}};
     9 int n,m,num;
    10 void DFS(int x,int y)
    11 {
    12     int a,b,i;
    13     vis[x][y]=1;
    14     num++;
    15     for(i=0;i<8;i++)
    16     {
    17         a=x+dir[i][0];
    18         b=y+dir[i][1];
    19         if(a>=0&&a<m&&b>=0&&b<n&&vis[a][b]==0&&map[a][b]=='.')
    20         {
    21             DFS(a,b);
    22         }
    23     }
    24 }
    25 int main()
    26 {
    27     int i,j,x,y;
    28     while(scanf("%d%d",&n,&m)!=EOF)
    29     {
    30         getchar();
    31         if(m==0&&n==0)
    32             break;
    33         memset(map,0,sizeof(map));
    34         memset(vis,0,sizeof(vis));
    35         for(i=0; i<m; i++)
    36         {
    37             scanf("%s",map[i]);
    38         }
    39         for(i=0; i<m; i++)
    40         {
    41             for(j=0; j<n; j++)
    42             {
    43                 if(map[i][j]=='@')///只有一个人
    44                 {
    45                    x=i;
    46                    y=j;
    47                 }
    48             }
    49         }
    50         num=0;
    51         DFS(x,y);
    52         printf("%d
    ",num);
    53     }
    54     return 0;
    55 }
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/wkfvawl/p/8906201.html
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