Vijos 1193 扫雷(状态压缩DP)

题目链接

思路很简单。。。但是细节啊。。。。WA了好多次。。。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <string>
 5 #include <ctime>
 6 #include <cstdlib>
 7 using namespace std;
 8 int p[10001];
 9 int dp1[10],dp2[10],o[3];
10 int main()
11 {
12     int i,j,n,k;
13     scanf("%d",&n);
14     for(i = 1;i <= n;i ++)
15     {
16         scanf("%d",&p[i]);
17     }
18     if(p[1] == 1)
19     {
20         dp1[1] = 1;
21         dp1[2] = 1;
22     }
23     else if(p[1] == 0)
24     dp1[0] = 1;
25     else if(p[1] == 2)
26     {
27         dp1[3] = 1;
28     }
29     if(n == 1)
30     {
31         if(p[1] == 1||p[1] == 0)
32         printf("%d\n",1);
33         else
34         printf("0\n");
35         return 0;
36     }
37     else if(n == 2)
38     {
39         if(p[1] == 0&&p[2] == 0)
40         printf("1\n");
41         else if(p[1] == 1&&p[2] == 1)
42         printf("2\n");
43         else if(p[1] == 2&&p[2] == 2)
44         printf("1\n");
45         else
46         printf("0\n");
47         return 0;
48     }
49     for(i = 3;i <= n;i ++)
50     {
51         for(j = 0;j < 4;j ++)
52         {
53             o[0] = o[1] = 0;
54             for(k = 0;k <= 1;k ++)
55             {
56                 if(j&(1<<k))
57                 o[k] ++;
58             }
59             if(o[0]+o[1] == p[i-1])
60             {
61                 dp2[o[1]] += dp1[j];
62             }
63             else if(o[0]+o[1]+1 == p[i-1])
64             {
65                 dp2[o[1]+2] += dp1[j];
66             }
67         }
68         for(j = 0;j < 4;j ++)
69         {
70             dp1[j] = dp2[j];
71             dp2[j] = 0;
72         }
73     }
74     if(p[n] == 2)
75     printf("%d\n",dp1[3]);
76     else if(p[n] == 0)
77     printf("%d\n",dp1[0]);
78     else if(p[n] == 1)
79     printf("%d\n",dp1[1]+dp1[2]);
80     else
81     printf("0\n");
82     return 0;
83 }

 

posted @ 2013-03-12 22:01  Naix_x  阅读(268)  评论(0)    收藏  举报