HDOJ 1312 DFS&BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337    Accepted Submission(s): 4591

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

 

广度搜索，需要用到队列：

 //bfs
 #include<stdio.h>
 int map[][];
 int queue[][];//队列
 int sx[]={,,,-};
 int sy[]={,,-,};
 int h,w;
 int ans;
 void bfs(int x,int y)
 {
     int rear=,front=;
     queue[rear][]=x;//enter queue
     queue[rear][]=y;
     rear++;
     while(front<rear)
     {
         for(int t=;t<;t++)
         {
             x=queue[front][]+sx[t];
             y=queue[front][]+sy[t];
             if(map[x][y]=='.'&&x>=&&x<=w&&y>=&&y<=h)
             {
                 queue[rear][]=x;
                 queue[rear][]=y;
                 map[x][y]='#';
                 rear++;
                 ans++;
             }
         }
         front++;
     }

 }
 int main()
 {
     int i,j;
     int x,y;
     while(scanf("%d%d",&h,&w)!=EOF)
     {
         if(!h&&!w)break;
         ans=;
         for(i=;i<=w;i++)
         {
             getchar();
             for(j=;j<=h;j++)
             {
                 scanf("%c",&map[i][j]);
                 if(map[i][j]=='@')
                     x=i,y=j;
             }
         }
         bfs(x,y);
         printf("%d\n",++ans);
     }
     return ;
 }

深度搜索：

 #include<stdio.h>
 int map[][];
 int t,n,m;
 int sx[]={,,-,};
 int sy[]={-,,,};
 void dfs(int h,int l)//深度搜索
 {
     int i,hx,hy;
     t++;
     map[h][l]='@';
     for(i=;i<;i++)
     {
         hx=h+sx[i];
         hy=l+sy[i];
         if(hx>=&&hx<=m&&hy>=&&hy<=n&&map[hx][hy]=='.')
             dfs(hx,hy);
     }
 }
 int main()
 {
     int a,b,x,y;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         if(n==&&m==)break;
         for(a=;a<=m;a++)
         {
             getchar();
             for(b=;b<=n;b++)
             {
                 scanf("%c",&map[a][b]);
                 if(map[a][b]=='@')
                 {
                     x=a;
                     y=b;
                 }
             }
         }
         t=;
         dfs(x,y);
         printf("%d\n",t);
     }
     return ;
 }