• HDU 1312 Red and Black (DFS)


    Red and Black

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9795    Accepted Submission(s): 6103


    Problem 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
     
    Source
     
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    搜索求出这个人能走多少黑色格子

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<cstring>
     4 #include<stdlib.h>
     5 #include<algorithm>
     6 using namespace std;
     7 const int MAXN=100;
     8 int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
     9 char map[MAXN][MAXN],s[MAXN];
    10 int vis[MAXN][MAXN];
    11 int ans,n,m;
    12 void DFS(int x,int y)
    13 {
    14     vis[x][y]=1;
    15     ans++;
    16     for(int i=0;i<4;i++)
    17     {
    18         int xx=x+dir[i][0];
    19         int yy=y+dir[i][1];
    20         if(0<=xx&&xx<n&&0<=yy&&yy<m&&!vis[xx][yy]&&map[xx][yy]!='#')
    21             DFS(xx,yy);
    22     }
    23 }
    24 int main()
    25 {
    26     //freopen("in.txt","r",stdin);
    27     int x,y;
    28     while(scanf("%d %d%*c",&m,&n)&&(n||m))
    29     {
    30         for(int i=0;i<n;i++)
    31         {
    32             for(int j=0;j<m;j++)
    33             {
    34                 scanf("%c",&map[i][j]);
    35                 if(map[i][j]=='@')
    36                 {
    37                     x=i;
    38                     y=j;
    39                 }
    40             }
    41             getchar();
    42         }
    43         memset(vis,0,sizeof(vis));
    44         ans=0;
    45         DFS(x,y);
    46         printf("%d
    ",ans);
    47     }
    48     return 0;
    49 }
    View Code
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  • 原文地址:https://www.cnblogs.com/clliff/p/3928204.html
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