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\n",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\n",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\n");

116     }

117     return  0 ;

118 }

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