6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
45
59
6
13
思路:dfs
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
using namespace std;
int c[4][2]={{0,-1},{0,1},{-1,0},{1,0}};//下一步的4种选择
int w,h,ans;
int x,y;
char s[22][22];
void dfs(int a,int b)
{
ans++;
s[a][b]=‘@‘;
int n,m;
for(int i=0;i<4;i++)
{
n=a+c[i][0];
m=b+c[i][1];
if(n<h&&m<w&&n>=0&&m>=0&&s[n][m]==‘.‘)
{
dfs(n,m);
}
}
}
int main()
{
while(cin>>w>>h&&(w!=0)&&(h!=0))
{
ans=0;
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
cin>>s[i][j];
if(s[i][j]==‘@‘)
{
x=i;
y=j;
}
}
}
dfs(x,y);
printf("%d\n",ans);
}
return 0;
}
原文:https://www.cnblogs.com/FantasticDoubleFish/p/14438102.html