hdu1312 赤裸BFS

Red and Black

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

45
59
6
13

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;
}