• CDZSC_2015寒假新人(4)——搜索 B


    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
     
     
     
    思路:就是从@开始上下左右4个方向把.全部找到,dfs就好
     
     
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 char str[25][25];
     6 int vis[25][25],w,h,sum;
     7 int x[]={-1,1,0,0};
     8 int y[]={0,0,-1,1};
     9 void dfs(int i,int j)
    10 {
    11     for(int k=0; k<4; k++)
    12     {
    13         if(i+x[k]<h&&i+x[k]>=0&&j+y[k]<w&&j+y[k]>=0&&vis[i+x[k]][j+y[k]]==0&&str[i+x[k]][j+y[k]]!='#')
    14         {
    15             sum++;
    16             vis[i+x[k]][j+y[k]]=1;
    17             dfs(i+x[k],j+y[k]);
    18         }
    19     }
    20 }
    21 int main()
    22 {
    23 #ifdef CDZSC_OFFLINE
    24     freopen("in.txt","r",stdin);
    25 #endif
    26     int i,j;
    27     while(scanf("%d%d",&w,&h)&&w!=0&&h!=0)
    28     {
    29         memset(vis,0,sizeof(vis));
    30         sum=1;
    31         for(i=0; i<h; i++)
    32         {
    33             scanf("%s",str[i]);
    34         }
    35         for(i=0; i<h; i++)
    36         {
    37             for(j=0; j<w; j++)
    38             {
    39                 if(str[i][j]=='@')
    40                 {
    41                     vis[i][j]=1;
    42                     dfs(i,j);
    43                 }
    44             }
    45         }
    46         printf("%d
    ",sum);
    47     }
    48     return 0;
    49 }
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  • 原文地址:https://www.cnblogs.com/Wing0624/p/4253923.html
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