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  • red and black(BFS)

    I - Red and Black

    Time Limit:1000MS     Memory Limit:32768KB     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) 

    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

    一道简单的广搜题,现在还是对这方面的编程太生疏,其实算法很简单但是自己不熟练所以感觉写的时候有些费力。

    AC代码:

     1 #include<bits/stdc++.h>
     2 #define N 25
     3 using namespace std;
     4 int vis[N][N];
     5 char mapn[N][N];
     6 int dir[4][2]={
     7     {1,0},
     8     {-1,0},
     9     {0,1},
    10     {0,-1},
    11 };
    12 int cur=0;
    13 int n,m;
    14 bool ok(int x,int y)
    15 {
    16     if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]=='#')
    17         return false;
    18     return true;
    19 }
    20 void dfs(int x,int y)
    21 {
    22     cur++;
    23     for(int i=0;i<4;i++)
    24     {
    25         int fx=x+dir[i][0];
    26         int fy=y+dir[i][1];
    27         if(ok(fx,fy))
    28         {
    29             vis[fx][fy]=1;
    30             dfs(fx,fy);
    31         }
    32     }
    33 }
    34 int main()
    35 {
    36     //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    37     while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
    38     {
    39         memset(vis,0,sizeof vis);
    40         int x,y;
    41         getchar();
    42         for(int i=0;i<m;i++)
    43         {
    44             scanf("%s",mapn[i]);
    45             //cout<<mapn[i]<<endl;
    46             for(int j=0;j<n;j++)
    47             {
    48                 if(mapn[i][j]=='@')
    49                 {
    50                     x=i;
    51                     y=j;
    52                 }
    53                 //cout<<mapn[i][j];
    54             }
    55             //cout<<endl;
    56         }
    57         cur=0;
    58         vis[x][y]=1;
    59         dfs(x,y);
    60         printf("%d
    ",cur);
    61     }
    62     return 0;
    63 }
    View Code
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/8638226.html
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