Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9795 Accepted Submission(s): 6103
搜索求出这个人能走多少黑色格子
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 const int MAXN=100; 8 int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; 9 char map[MAXN][MAXN],s[MAXN]; 10 int vis[MAXN][MAXN]; 11 int ans,n,m; 12 void DFS(int x,int y) 13 { 14 vis[x][y]=1; 15 ans++; 16 for(int i=0;i<4;i++) 17 { 18 int xx=x+dir[i][0]; 19 int yy=y+dir[i][1]; 20 if(0<=xx&&xx<n&&0<=yy&&yy<m&&!vis[xx][yy]&&map[xx][yy]!=‘#‘) 21 DFS(xx,yy); 22 } 23 } 24 int main() 25 { 26 //freopen("in.txt","r",stdin); 27 int x,y; 28 while(scanf("%d %d%*c",&m,&n)&&(n||m)) 29 { 30 for(int i=0;i<n;i++) 31 { 32 for(int j=0;j<m;j++) 33 { 34 scanf("%c",&map[i][j]); 35 if(map[i][j]==‘@‘) 36 { 37 x=i; 38 y=j; 39 } 40 } 41 getchar(); 42 } 43 memset(vis,0,sizeof(vis)); 44 ans=0; 45 DFS(x,y); 46 printf("%d\n",ans); 47 } 48 return 0; 49 }
HDU 1312 Red and Black (搜索),布布扣,bubuko.com
原文:http://www.cnblogs.com/clliff/p/3928204.html