簡(jiǎn)單DFS

注意邊界就行了。

// #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef pair<int,int> pii; #define pb(a) push_back(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,rvoid debug() { #ifdef ONLINE_JUDGE #elsefreopen("d:\\in.txt","r",stdin);freopen("d:\\out1.txt","w",stdout); #endif } char getch() {char ch;while((ch=getchar())!=EOF){if(ch!=' '&&ch!='\n')return ch;}return EOF; } char da[25][25]; int n,m; int num; int dx[]={-1,0,1,0}; int dy[]={0,1,0,-1}; int vis[25][25]; int dfs(int x,int y) {vis[x][y]=1;num++;for(int d=0;d<4;d++){int nx=x+dx[d],ny=y+dy[d];if(da[nx][ny]=='.'&&!vis[nx][ny]&&nx>=1&&nx<=n&&ny>=1&&ny<=m)dfs(nx,ny);}return 0; } int main() {while(scanf("%d%d",&m,&n)!=EOF&&(n||m)){for(int i=1; i<=n; i++)scanf("%s",da[i]+1);num=0;int bx,by;for(int i=1; i<=n; i++)for(int j=1; j<=m; j++){if(da[i][j]=='@'){bx=i;by=j;break;}}memset(vis,0,sizeof(vis));dfs(bx,by);printf("%d\n",num);}return 0; } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/BMan/p/3238104.html

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