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HDU 1312 Red and Black (搜索)

2024-07-18 20:36:43   226人阅读

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
Asia 2004, Ehime (Japan), Japan Domestic
 

 

Recommend
Eddy
 

 

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\n",ans);47     }48     return 0;49 }
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