hdu1312 赤裸BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3893    Accepted Submission(s): 2505

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

4559613
一个纯的BFS    赤裸裸的BFS   一开始wa在按照自己的斯洛没有考虑到类似这种####@数据 也就是ans=1的时候的情狂
#include<stdio.h>#include<string.h>int a[22][22],w,h;int c[4][2]={1,0,0,1,-1,0,0,-1};int s_x,s_y;struct haha{ int x; int y;}q[1000000];int bfs(){        int head=0,tail=1,ans=0,i,x1,y1;  q[1].x=s_x;q[1].y=s_y;  while(head<tail)  {            ++head;   for(i=0;i<4;i++)     {                x1=q[head].x+c[i][0];    y1=q[head].y+c[i][1];    if(a[x1][y1]!=0)//&&x1>0&&x1<=h&&y1>0&&y1<=w    {     //printf("a[%d][%d]=%d",x1,y1,a[x1][y1]);     ans++;     a[x1][y1]=0;     q[++tail].x=x1;     q[tail].y=y1;    }   }  }  if(ans==0) return 1;//一开始把这里忘了 由于每次找到能走的则加加   else//最后一次加加相当于多加了一次  我把他当做原始位置的那一个了 但是忘记了考虑   return ans;//当只有原始位置的时候  它不能进入循环 ans=0 但是原始位置的那次没加上}int main(){        int i,j,ans;  char ch;  while(scanf("%d %d",&w,&h)==2)  {   getchar();   if(w<=0||h<=0) return 0;   memset(a,0,sizeof(a));   ans=1;   for(i=1;i<=h;i++)   {    for(j=1;j<=w;j++)    { ch=getchar();     if(ch=='@'){s_x=i;s_y=j;a[i][j]=1;}//printf("s_x=%d,s_y=%d\n",s_x,s_y);     if(ch=='.') a[i][j]=1;     if(ch=='#') a[i][j]=0;    }    getchar();   }            ans=bfs();   printf("%d\n",ans);  }  return 0;}