USACO 2.4 The Tamworth Two(乱搞）

这个和金华的小模拟很相似，题目中啥也没说，想枚举100000秒，试一下水的。结果就水过了，1Y。

 1 /*

 2   ID: cuizhe

 3   LANG: C++

 4   TASK: ttwo

 5  */

 6 #include <iostream>

 7 #include <cstdio>

 8 #include <cstring>

 9 #include <cmath>

10 #include <algorithm>

11 using namespace std;

12 int x[4] = {-1,0,1,0};

13 int y[4] = {0,1,0,-1};

14 char str[11][11];

15 int judge(int r,int c,int d)

16 {

17     if(r+x[d] > 9||r+x[d] < 0)

18     return 0;

19     if(c+y[d] > 9||c+y[d] < 0)

20     return 0;

21     if(str[r+x[d]][c+y[d]] == '*')

22     return 0;

23     return 1;

24 }

25 int main()

26 {

27     int i,j,r1,c1,r2,c2,d1,d2;

28     freopen("ttwo.in","r",stdin);

29     freopen("ttwo.out","w",stdout);

30     for(i = 0;i <= 9;i ++)

31     scanf("%s",str[i]);

32     d1 = d2 = 0;

33     for(i = 0;i <= 9;i ++)

34     {

35         for(j = 0;j <= 9;j ++)

36         {

37             if(str[i][j] == 'F')

38             {

39                 r1 = i;

40                 c1 = j;

41             }

42             if(str[i][j] == 'C')

43             {

44                 r2 = i;

45                 c2 = j;

46             }

47         }

48     }

49     for(i = 1;i <= 100000;i ++)

50     {

51         if(judge(r1,c1,d1))

52         {

53             r1 = r1+x[d1];

54             c1 = c1+y[d1];

55         }

56         else

57         d1 = (d1+1)%4;

58         if(judge(r2,c2,d2))

59         {

60             r2 = r2+x[d2];

61             c2 = c2+y[d2];

62         }

63         else

64         d2 = (d2+1)%4;

65         if(r1 == r2&&c1 == c2)

66         {

67             printf("%d\n",i);

68             break;

69         }

70     }

71     if(i == 100001)

72     printf("0\n");

73     return 0;

74 }