POJ2632Crashing Robots

做模拟题做的我直接睡着了，题并不难，就是一个细心的问题，有一些细节问题注意了就差不多了，代码写的精美的一般找错误也好找一些，应该学着些好看的代码
  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 using namespace std ;
  5 int len,high;
  6 int a,b,j ;
  7 struct node
  8 {
  9     int x;
 10     int y ;
 11     int dire;
 12 } s[10001];
 13 int judge(int k)
 14 {
 15     int i;
 16     if(s[k].x > len||s[k].x< 1||s[k].y>high||s[k].y<1)
 17     {
 18         printf("Robot %d crashes into the wall
",k);
 19         return 0;
 20     }
 21     for(i = 1 ; i <= a ; i++)
 22     {
 23         if(i == k)
 24             continue;
 25         if(s[i].x == s[k].x&&s[i].y == s[k].y)
 26         {
 27             printf("Robot %d crashes into robot %d
",k,i);
 28             return 0 ;
 29         }
 30     }
 31     return 1 ;
 32 }
 33 int main()
 34 {
 35     int n ;
 36     cin>>n;
 37     for(int i = 1 ; i <= n ; i++)
 38     {
 39         cin>>len>>high;
 40         cin>>a>>b ;
 41         char dire ;
 42         for(j = 1 ; j <= a ; j++)
 43         {
 44             cin>>s[j].x>>s[j].y>>dire;
 45             if(dire == 'N')
 46                 s[j].dire = 0 ;
 47             if(dire == 'W')
 48                 s[j].dire = 1 ;
 49             if(dire == 'S')
 50                 s[j].dire = 2 ;
 51             if(dire == 'E')
 52                 s[j].dire = 3 ;
 53         }
 54         int num,repeat,flag = 1 ;
 55         char order ;
 56         for(j = 1 ; j <= b ; j++)
 57         {
 58             cin>>num>>order>>repeat ;
 59             for(int h = 1 ; h <= repeat ; h++ )//把这个放在外边是为了底下的左右指令时比较好处理
 60             {
 61                 if(order == 'F')
 62                 {
 63                     if(s[num].dire == 3)
 64                     {
 65                         s[num].x++ ;
 66                         if(!judge(num))
 67                         {
 68                             flag = 0 ;
 69                             break ;
 70                         }
 71                     }
 72                     if(s[num].dire == 1)
 73                     {
 74                         s[num].x--;
 75                         if(!judge(num))
 76                         {
 77                             flag = 0 ;
 78                             break ;
 79                         }
 80                     }
 81                     if(s[num].dire == 0)
 82                     {
 83                         s[num].y++ ;
 84                         if(!judge(num))
 85                         {
 86                             flag = 0 ;
 87                             break ;
 88                         }
 89                     }
 90                     if(s[num].dire == 2)
 91                     {
 92                         s[num].y--;
 93                         if(!judge(num))
 94                         {
 95                             flag = 0 ;
 96                             break ;
 97                         }
 98                     }
 99                 }
100 
101                 if(order == 'L')
102                     s[num].dire = (1+s[num].dire)%4 ;
103                 if(order == 'R')
104                     s[num].dire = (s[num].dire-1+4)%4;
105             }
106             if(flag == 0)
107                 break ;
108         }
109         if(j < b)
110             for(++j ; j <= b ; j++)
111             {
112                 cin>>num>>order>>repeat ;
113             }
114         if(flag == 1)
115             printf("OK
");
116     }
117     return  0 ;
118 }
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