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 }

 

 

posted @ 2012-11-14 17:13  Naix_x  阅读(224)  评论(0)  编辑  收藏  举报