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

....#.
.....#
......
......
......
......
......
#@...#
.#..#.

.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........

..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..

..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
 

Sample Output

45

59

6

13

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

Recommend

Eddy   |   We have carefully selected several similar problems for you:  1372 1242 1240 1072 1258 

---

 

　　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：

 #include <iostream>
 using namespace std;
 int cnt;
 char a[][];
 int n,m;
 int dx[] = {,,,-};    //方向
 int dy[] = {,,-,};
 bool judge(int x,int y)
 {
     if(x< || x>n || y< || y>m)
         return ;
     if(a[x][y]=='#')
         return ;
     return ;
 }
 void dfs(int cx,int cy)
 {
     cnt++;
     a[cx][cy] = '#';
     int i;
     for(i=;i<;i++){
         int nx = cx + dx[i];
         int ny = cy + dy[i];
         if(judge(nx,ny))
             continue;
         //可以走
         dfs(nx,ny);
     }
 }
 int main()
 {
     while(cin>>m>>n){
         if(n== && m==) break;
         int i,j;
         int x,y;
         for(i=;i<=n;i++)
             for(j=;j<=m;j++){
                 cin>>a[i][j];
                 if(a[i][j]=='@')    //记录开始的位置
                     x=i,y=j;
             }
         cnt = ;
         dfs(x,y);
         cout<<cnt<<endl;
     }
     return ;
 }

 　　这种方法直接返回结果，两种不同的写法，代码二：

 #include <iostream>
 using namespace std;
 char a[][];
 int n,m;
 int dx[] = {,,,-};    //方向
 int dy[] = {,,-,};
 bool judge(int x,int y)
 {
     if(x< || x>n || y< || y>m)
         return ;
     if(a[x][y]=='#')
         return ;
     return ;
 }
 int dfs(int cx,int cy)
 {
     int i,sum=;
     a[cx][cy] = '#';    //走过的这一步覆盖
     for(i=;i<;i++){
         int nx = cx + dx[i];
         int ny = cy + dy[i];
         if(judge(nx,ny))
             continue;
         //可以走
         sum+=dfs(nx,ny);
     }
     return sum==?:sum+;
 }
 int main()
 {
     while(cin>>m>>n){
         if(n== && m==) break;
         int i,j;
         int x,y;
         for(i=;i<=n;i++)
             for(j=;j<=m;j++){
                 cin>>a[i][j];
                 if(a[i][j]=='@')    //记录开始的位置
                     x=i,y=j;
             }
         cout<<dfs(x,y)<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013