Red and Black
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) The end of the input is indicated by a line consisting of two zeros.
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
http://acm.pku.edu.cn/JudgeOnline/problem?id=1979
简单的bfs，最不地道的是给的例子就一次换行，judge时按每一个结果都换，害的我提交几次，都wa
#include<stdio.h>
typedef struct Q{
int x,y;
}Queue;
Queue result[1000];
int tail,head;
int W,H;
void bfs(int input_rec_map[][30],int x,int y){
int dir[2] ={1,-1};
int i;
if(head == tail){
return ;
}
for(i = 0; i < 2;++i){
if((x+dir[i] > -1) && (x+dir[i] < W) && input_rec_map[y][x+ dir[i]] ){
input_rec_map[y][x+ dir[i]] = 0;
result[tail].x = x + dir[i];
result[tail].y = y;
++tail;
}
}
for(i = 0; i < 2;++i){
if((y + dir[i] > -1) && (y + dir[i] < H) &&(input_rec_map[y+dir[i]][x])){
input_rec_map[y+dir[i]][x] = 0;
result[tail].x = x;
result[tail].y = y + dir[i];
++tail;
}
}
++head;
bfs(input_rec_map,result[head].x,result[head].y);
}
int main(){
char input_rec[30][30];
int input_rec_map[30][30];
int i,j;
//int than_one = 0;
//FILE *fp;
//fp = freopen("in.txt","r",stdin);
while(~scanf("%d%d",&W,&H),W||H){
for(i = 0;i < H; ++i){
scanf("%s",input_rec[i]);
}
for(i = 0; i < H; ++i){
for(j = 0;j<W;++j){
if(input_rec[i][j] == '.'){
input_rec_map[i][j] = 1;
}else if(input_rec[i][j] == '#'){
input_rec_map[i][j] = 0;
}else{
result[0].x = j;
result[0].y = i;
input_rec_map[i][j] = 0;
head = 0;
tail = 1;
}
}
}
bfs(input_rec_map,result[0].x,result[0].y);
//if(than_one){
// printf(" %d",head);
//}else{
printf("%d\n",head);
//}
//than_one = 1;
}
//printf("\n");
return 0;
}