hdu 1312:Red and Black（DFS搜索，入门题）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8435    Accepted Submission(s): 5248

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

 

 

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　　DFS搜索，入门题。

　　规定地图中有可通行的位置，也有不可通行的位置，已知起点，求一个连通分量。说白了就是求一个点的与它相连的部分。在这道题里输出相连的位置的数目。

　　思路是从起点开始，遍历每一个到达的点的四个方向，到达一个位置就将这个位置的字符变成不可走的'#'，并且计数+1。其实就是计数将可走变成不可走的操作进行了多少次。

　　代码一：

 1 #include <iostream>

 2 using namespace std;  3 int cnt;  4 char a[22][22];  5 int n,m;  6 int dx[4] = {0,1,0,-1};    //方向

 7 int dy[4] = {1,0,-1,0};  8 bool judge(int x,int y)  9 { 10     if(x<1 || x>n || y<1 || y>m) 11         return 1; 12     if(a[x][y]=='#') 13         return 1; 14     return 0; 15 } 16 void dfs(int cx,int cy) 17 { 18     cnt++; 19     a[cx][cy] = '#'; 20     int i; 21     for(i=0;i<4;i++){ 22         int nx = cx + dx[i]; 23         int ny = cy + dy[i]; 24         if(judge(nx,ny)) 25             continue; 26         //可以走

27  dfs(nx,ny); 28  } 29 } 30 int main() 31 { 32     while(cin>>m>>n){ 33         if(n==0 && m==0) break; 34         int i,j; 35         int x,y; 36         for(i=1;i<=n;i++) 37             for(j=1;j<=m;j++){ 38                 cin>>a[i][j]; 39                 if(a[i][j]=='@')    //记录开始的位置

40                     x=i,y=j; 41  } 42         cnt = 0; 43  dfs(x,y); 44         cout<<cnt<<endl; 45  } 46     return 0; 47 }

 　　这种方法直接返回结果，两种不同的写法，代码二：

 1 #include <iostream>

 2 using namespace std;  3 char a[22][22];  4 int n,m;  5 int dx[4] = {0,1,0,-1};    //方向

 6 int dy[4] = {1,0,-1,0};  7 bool judge(int x,int y)  8 {  9     if(x<1 || x>n || y<1 || y>m) 10         return 1; 11     if(a[x][y]=='#') 12         return 1; 13     return 0; 14 } 15 int dfs(int cx,int cy) 16 { 17     int i,sum=0; 18     a[cx][cy] = '#';    //走过的这一步覆盖

19     for(i=0;i<4;i++){ 20         int nx = cx + dx[i]; 21         int ny = cy + dy[i]; 22         if(judge(nx,ny)) 23             continue; 24         //可以走

25         sum+=dfs(nx,ny); 26  } 27     return sum==0?1:sum+1; 28 } 29 int main() 30 { 31     while(cin>>m>>n){ 32         if(n==0 && m==0) break; 33         int i,j; 34         int x,y; 35         for(i=1;i<=n;i++) 36             for(j=1;j<=m;j++){ 37                 cin>>a[i][j]; 38                 if(a[i][j]=='@')    //记录开始的位置

39                     x=i,y=j; 40  } 41         cout<<dfs(x,y)<<endl; 42  } 43     return 0; 44 }

 

Freecode : www.cnblogs.com/yym2013