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
题意:一个人在一个H*W矩形房间中,当前位置为 @ 点。他只能沿上下左右4个方向走,且只能走 . 点,不能走 # 点,已走过的 . 点可重复走,问他最多可走多少 . 点。
由于已走过的 . 点可重复走,故可考虑无回溯DFS,将走过的 . 点标记为 #,这样可避免设置vis数组。最终结果即走过的 . 数目。
#include<string.h>
#include<stdio.h>
#include<algorithm>
#define N 20
int n,m;
char map[N][N];
bool have[N][N];
int count,trueRow,trueCol;
void bfs(int row, int col){
if(row < n && row >= 0 && col < m && col>=0){//越界
if((map[row][col] == '.' || map[row][col] == '@' ) && have[row][col] == false){
have[row][col] = true;//开始的时候没把’@‘加上,盲区了吧
count++; //计数
bfs(row-1,col);//上下左右遍历
bfs(row,col+1);
bfs(row+1,col);
bfs(row,col-1);
}
}
}
int main()
{
int i,j;
scanf("%d%d",&m,&n);
getchar(); //输入n之后的换行符
memset(have,false,sizeof(have));
for(i=0; i<n; i++){
for(j=0; j<m; j++){
scanf("%c", &map[i][j]);
if(map[i][j] == '@'){//确定开始位置
trueRow = i;
trueCol = j;
}
}
getchar();//输入一行之后的换行符
}
count = 0;//全局的能联通的个数
bfs(trueRow, trueCol);//宽度优先遍历
printf("%d\n",count);
return 0;
}