速刷一道DFS
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.
@是起点，#不能走，'.'是正常道路，问从起点能到达多少格子
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
刷道普通的DFS，复习模板
/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=;
char mp[mxn][mxn];
int ans;
int W,H;
void dfs(int x,int y){
if(x< || x>H || y< || y>W)return;
if(mp[x][y]=='#')return;
ans++;
mp[x][y]='#';
dfs(x+,y);
dfs(x,y+);
dfs(x-,y);
dfs(x,y-);
return;
}
int main(){
while(scanf("%d%d",&W,&H) && W && H){
memset(mp,' ',sizeof(mp));
ans=;
int i,j;
for(i=;i<=H;i++){
scanf("%s",mp[i]+);
}
int sx,sy;
bool flag=;
for(i=;i<=H;i++){
for(j=;j<=W;j++)
if(mp[i][j]=='@'){
sx=i;sy=j;
flag=;
break;
}
if(flag)break;
}
dfs(sx,sy);
printf("%d\n",ans);
}
return ;
}