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