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Peter's Hobby

2024-07-16 03:49:17   224人阅读

Problem Description
Recently, Peter likes to measure the humidity of leaves. He recorded a leaf humidity every day. There are four types of leaves wetness: Dry , Dryish , Damp and Soggy. As we know, the humidity of leaves is affected by the weather. And there are only three kinds of weather: Sunny, Cloudy and Rainy.For example, under Sunny conditions, the possibility of leaves are dry is 0.6.
Give you the possibility list of weather to the humidity of leaves.


The weather today is affected by the weather yesterday. For example, if yesterday is Sunny, the possibility of today cloudy is 0.375.
The relationship between weather today and weather yesterday is following by table:


Now,Peter has some recodes of the humidity of leaves in N days.And we know the weather conditons on the first day : the probability of sunny is 0.63,the probability of cloudy is 0.17,the probability of rainny is 0.2.Could you know the weathers of these days most probably like in order?
 

 

Input
The first line is T, means the number of cases, then the followings are T cases. for each case:
The first line is a integer n(n<=50),means the number of days, and the next n lines, each line is a string shows the humidity of leaves (Dry, Dryish, Damp, Soggy)
 

 

Output
For each test case, print the case number on its own line. Then is the most possible weather sequence.( We guarantee that the data has a unique solution)
 

 

Sample Input
1
3
Dry
Damp
Soggy
 

 

Sample Output
Case #1:
Sunny
Cloudy
Rainy
Hint
Log is useful.
概率dp  time :0ms
#include"iostream"#include"cstdio"#include"map"#include"cstring"#include"string"#include"algorithm"#include"vector"using namespace std;const int ms=51;double p1[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};double p2[3][3]= {0.5,0.375,0.125,0.25,0.125,0.625,0.25,0.375,0.375};double dp[ms][3];int pre[ms][3];map<int,string> m1;map<string,int> m2;string str;void init(){    m2.insert(make_pair("Dry",0));    m2.insert(make_pair("Dryish",1));    m2.insert(make_pair("Damp",2));    m2.insert(make_pair("Soggy",3));    m1.insert(make_pair(0,"Sunny"));    m1.insert(make_pair(1,"Cloudy"));    m1.insert(make_pair(2,"Rainy"));}int main(){    int ncase,T=0;    scanf("%d",&ncase);    init();    while(ncase--)    {        printf("Case #%d:\n",++T);        int n,i,j,k;        scanf("%d",&n);        cin>>str;        for(i=0;i<=n;i++)            for(j=0;j<3;j++)                dp[i][j]=0;        int lab=m2[str];        memset(pre,0,sizeof(pre));        dp[1][0]=0.63*p1[0][lab];        dp[1][1]=0.17*p1[1][lab];        dp[1][2]=0.2*p1[2][lab];        for(i=2;i<=n;i++)        {            cin>>str;            lab=m2[str];            for(j=0;j<3;j++)                for(k=0;k<3;k++)                {                    double pp=dp[i-1][k]*p2[k][j]*p1[j][lab];                    if(pp>dp[i][j])                    {                        dp[i][j]=pp;                        pre[i][j]=k;                    }                }        }        vector<int> ans;        double mi=0;        int po;        for(i=0;i<3;i++)            if(dp[n][i]>mi)            {                mi=dp[n][i];                po=i;            }        ans.push_back(po);        int now=n;        while(now!=1)        {            po=pre[now][po];            ans.push_back(po);            now--;        }        for(i=n-1;i>=0;i--)        {            //printf("%s\n",m1[ans[i]]);            cout<<m1[ans[i]]<<endl;        }    }    return 0;}

 


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