这题简直把我坑死了 所有的坑都被我中了

题意：

思路：bfs or 模拟 模拟似乎没有什么坑 但是bfs真的是坑 

AC代码：

1 #include "iostream" 2 #include "string.h" 3 #include "stack" 4 #include "queue" 5 #include "string" 6 #include "vector" 7 #include "set" 8 #include "map" 9 #include "algorithm"10 #include "stdio.h"11 #include "math.h"12 #define ll long long13 #define ull unsigned ll14 #define mem(a) memset(a,0,sizeof(a))15 #define bug cout<<"UUUUUUUUUUUUU\n";16 using namespace std;17 struct Node{18     int xx,yy;19 };20 int d[8][2]{
   {
   1,1},{
   1,0},{
   0,1},{
   1,-1},{-1,0},{-1,-1},{
   0,-1},{-1,1}};21 int flag;22 char m[10][10];23 void Bfs(int x,int y){24     Node now,next;25     int vis[10][10];26     mem(vis);27     queue
     
       Q;28     while(!Q.empty()) Q.pop();29     now.xx=x,now.yy=y;30     Q.push(now);31     vis[x][y]=1;32     while(!Q.empty()){33         now=Q.front();34         Q.pop();35         for(int i=0; i<8; i++){36             next.xx=now.xx+d[i][0];37             next.yy=now.yy+d[i][1];38             int s=1;39             int f=0;40             if(m[now.xx][now.yy]=='x') f=1;41             if(!vis[next.xx][next.yy]&&(next.xx>0&&next.xx<5&&next.yy>0&&next.yy<5&&(m[next.xx][next.yy]=='.'||m[next.xx][next.yy]=='x'))){42                 Q.push(next);43                 vis[next.xx][next.yy]=1;44             }45             while(next.xx>0&&next.xx<5&&next.yy>0&&next.yy<5&&(m[next.xx][next.yy]=='.'||m[next.xx][next.yy]=='x')){46                 s++;47                 if(m[next.xx][next.yy]=='x') f++; //printf("%d %d\n%d %d %d\n",s,f,next.xx,next.yy,i);48                 if(s==3&&f>=2) {printf("YES\n");flag=1;return;}49                 next.xx+=d[i][0];50                 next.yy+=d[i][1];51             }52         }53     }54 }55 int main(){56     mem(m);57     int x1,y1;58     for(int i=1; i<=4; ++i)59         for(int j=1; j<=4; ++j)60             cin>>m[i][j];61     for(int i=1; i<5; ++i){62         for(int j=1; j<5; ++j){63             if(!flag&&(m[i][j]=='.'||m[i][j]=='x')) Bfs(i,j);64         }65     }66     if(!flag) printf("NO\n");67     return 0;68 }69 /*70 o.x.71 o...72 .x..73 ooxx74 75 x.ox76 ox..77 x.o.78 oo.x79 80 xoxx81 ..x.82 o.oo83 x.o.84 */