HDU 4865 Peter's Hobby --概率DP

题意：第i天的天气会一定概率地影响第i+1天的天气，也会一定概率地影响这一天的湿度.概率在表中给出。给出n天的湿度，推测概率最大的这n天的天气。

分析：这是引自机器学习中隐马尔科夫模型的入门模型，其实在这里直接DP就可以了

定义：dp[i][j]为第i天天气为j（0,1,2分别表示三个天气）的概率，path[i][j]记录路径,path[i][j] = k 意思是前一天天气为k时，这一天有最大的概率是天气j。

做一个三重循环，对于每天，枚举今天的天气，再在里面枚举昨天的天气，则有：

dp[i][j] = max(dp[i-1][k]*yto[k][j]*wtoh[j][humi[i]])

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#define eps 1e-4
using namespace std;

string weather[4] = {"Sunny","Cloudy","Rainy"};
double yto[3][3]={{0.5,0.375,0.125},{0.25,0.125,0.625},{0.25,0.375,0.375}};
double wtoh[3][4]={{0.6,0.2,0.15,0.05},{0.25,0.3,0.2,0.25},{0.05,0.10,0.35,0.50}};
int humi[55],path[55][3],ans[55];
double dp[55][4];

int gethum(string ss)
{
    if(ss == "Dry")
        return 0;
    else if(ss == "Dryish")
        return 1;
    else if(ss == "Damp")
        return 2;
    else
        return 3;
}

int main()
{
    int t,cs = 1,i,j,n,k;
    string ss;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            cin>>ss;
            int hum = gethum(ss);
            humi[i] = hum;
        }
        for(i=0;i<=n;i++)
            for(j=0;j<3;j++)
                dp[i][j] = 0.0;
        memset(path,0,sizeof(path));
        dp[0][0] = 0.63*wtoh[0][humi[0]];
        dp[0][1] = 0.17*wtoh[1][humi[0]];
        dp[0][2] = 0.20*wtoh[2][humi[0]];
        for(i=1;i<n;i++)
        {
            for(j=0;j<3;j++)  //today's weather
            {
                for(k=0;k<3;k++)  //yesterday's weather
                {
                    double P = dp[i-1][k]*yto[k][j]*wtoh[j][humi[i]];
                    if(P > dp[i][j])
                    {
                        dp[i][j] = P;
                        path[i][j] = k;
                    }
                }
            }
        }
        int now = 0;
        for(i=0;i<3;i++)
        {
            if(dp[n-1][i] > dp[n-1][now])
                now = i;
        }
        ans[n-1] = now;
        for(i=n-2;i>=0;i--)
        {
            now = path[i+1][now];
            ans[i] = now;
        }
        printf("Case #%d:
",cs++);
        for(i=0;i<n;i++)
            cout<<weather[ans[i]]<<endl;
    }
    return 0;
}