hdu 4865 Peter's Hobby(2014 多校联合第一场 E)

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 char yezi[5][10]= {"Dry","Dryish","Damp","Soggy"};
 5 char wea[4][10]= {"Sunny","Cloudy","Rainy"};
 6 double yeP[3][4]= {0.6,0.2,0.15,0.05,
 7                    0.25,0.3,0.2,0.25,
 8                    0.05,0.10,0.35,0.50
 9                   };
10 double weaP[3][3]= {0.5,0.375,0.125,
11                     0.25,0.125,0.625,
12                     0.25,0.375,0.375
13                    };
14 double init[3]= {0.63,0.17,0.2};
15 double dp[51][3],pre[51][3];
16 int n;
17 double maxp;
18 int Pos[51];
19 void solve()
20 {
21     for(int i=0; i<3; i++)
22     {
23         dp[1][i]=log(init[i])+log(yeP[i][Pos[1]]);
24                  //第1天天气为i的概率=初始的天气为i的概率*（第一天叶子为Pos[1]的的状态下天气为i的概率）
25     }
26     for(int i=2; i<=n; i++) //
27     {
28         for(int j=0; j<3; j++) //天气
29         {
30             double maxp=-1e8;
31             int pos=0;//记录天气
32             for(int k=0; k<3; k++) //第i-1天天气为k
33             {
34                 double temp=dp[i-1][k]+log(weaP[k][j])+log(yeP[j][Pos[i]]);
35                 //昨天天气为k的概率*昨天天气为k今天天气为j的概率*叶子状态为Pos[i]时天气为j的概率
36                 if(temp>maxp)
37                 {
38                     maxp=temp;
39                     pos=k;//记录最有可能的天气状况为k
40                 }
41             }
42             dp[i][j]=maxp;//第i天天气状况为j的概率
43             pre[i][j]=pos; //表示第i天天气最可能为j的前一天天气
44         }
45     }
46 }
47 int main()
48 {
49     int k=0;
50     int T;
51     char yeC[10];
52     scanf("%d",&T);
53     while(T--)
54     {
55         k++;
56         printf("Case #%d:\n",k);
57         scanf("%d",&n);
58         for(int i=1; i<=n; i++)
59         {
60             scanf("%s",yeC);
61             for(int j=0; j<4; j++)
62                 if(strcmp(yeC,yezi[j])==0)
63                 {
64                     Pos[i]=j;
65                     break;
66                 }
67         }
68         solve();
69         double maxp=-1e8;
70         int ans[51];
71         for(int i=0; i<3; i++)//第n天最有可能的天气
72             if(dp[n][i]>maxp)
73             {
74                 maxp=dp[n][i];
75                 ans[n]=i;
76             }
77          for(int i=n-1;i>=1;i--)
78             ans[i]=pre[i+1][ans[i+1]];  //由最后一天往前找每天的天气状况记录在ans 中
79         for(int i=1;i<=n;i++)
80             printf("%s\n",wea[ans[i]]);
81     }
82     return 0;
83 }