HDU4414-DFS

给一个图，寻找十字交叉的个数，十字交叉应为两个大于3的奇数交叉与正中央。图的大小很小。
使用DFS搜八连块，之后按照规则筛选出符合条件的交叉。
我的筛选规则有点蠢，先将点排序，再通过三段for循环判断。
  1 #include <algorithm>
  2 #include <cstring>
  3 #include <ctype.h>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <vector>
  7 #include <string>
  8 #include <queue>
  9 #include <stack>
 10 #include <cmath>
 11 #include <set>
 12 #include <map>
 13 
 14 using namespace std;
 15 
 16 struct Node{
 17     int x;
 18     int y;
 19     bool operator < (const struct Node &b) const 
 20     {
 21         if(x <= b.x)
 22         {
 23             if(x == b.x) return y <= b.y;
 24             else return true;
 25         }
 26         return false;
 27     }
 28 }cross[5000];
 29 
 30 int N,M,T;
 31 char G[100100];
 32 int vis[100100];
 33 int dx[] = {1,-1,0,0},dy[] = {0,0,1,-1};
 34 int P;
 35 
 36 void dfs(int u)
 37 {
 38     vis[u] = true;    
 39     cross[P].x = u%N;cross[P].y = u/N;
 40     P++;
 41     for(int i=0;i<4;i++)
 42     {
 43         int x = u%N + dx[i],y = u/N + dy[i];
 44         if(x >=0 && x < N && y >= 0 && y < N &&!vis[y*N+x] && G[y*N+x] == '#')
 45         {
 46             dfs(x+y*N);
 47         }
 48     }
 49 }
 50 
 51 int main()
 52 {
 53     while(scanf("%d ",&N) && N)
 54     {
 55         char s[100];
 56         for(int i=0;i<N;i++)
 57         {
 58             scanf("%s",s);
 59             for(int j=0;j<N;j++)
 60             {
 61                 G[i*N+j] = s[j];
 62             }
 63         }
 64 
 65         memset(vis,0,sizeof vis);
 66         int ans = 0;
 67         for(int i=0;i<N;i++)
 68         {
 69             for(int j=0;j<N;j++)
 70             {
 71                 if(G[i*N+j] != '#' || vis[i*N+j]) continue;
 72                 P = 0;
 73                 dfs(i*N+j);
 74             //    printf("P=%d
",P);
 75                 if(P % 2 == 0 || P < 5) continue;
 76                 else
 77                 {
 78                     //printf("check
");
 79                     sort(cross,cross+P);
 80                     /*
 81                     for(int i=0;i<P;i++)
 82                     {
 83                         printf("%d:(%d,%d)
",i,cross[i].x,cross[i].y);
 84                     }
 85                     */
 86                     int len = (P+1)/2,ok = 1,step = 0;
 87                     for(int i=0;i<(len-1)/2;i++)
 88                     {
 89                         if(cross[i].x != cross[0].x+i || cross[i].y != cross[0].y) {ok = 0;break;}
 90                     }
 91                     step += (len-1)/2;
 92                     for(int i=step;i < len+step;i++)
 93                     {
 94                         if(cross[i].y != (cross[step].y+i-step)
 95                         || cross[i].x != cross[step].x 
 96                         || cross[step].x != cross[step-1].x+1
 97                         || cross[step].y != cross[step-1].y-(len-1)/2) 
 98                         {ok = 0;break;}
 99                     }
100                     step += len;
101                     for(int i=step;i< step + (len-1)/2;i++)
102                     {
103                         if(cross[i].x != (cross[step].x+i-step) 
104                         || cross[i].y != cross[step].y 
105                         || cross[step].y != cross[0].y) 
106                         {ok = 0;break;}
107                     }
108                     if(ok) ans++;
109                 }
110 
111             }
112         }
113         printf("%d
",ans);
114     }
115 }