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:\n",cs++);

        for(i=0;i<n;i++)

            cout<<weather[ans[i]]<<endl;

    }

    return 0;

}

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