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HDU 4865 Peter's Hobby

2024-07-18 07:26:12   226人阅读

Peter‘s Hobby

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 647    Accepted Submission(s): 273

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.
 
Author
FZU
 
Source
2014 Multi-University Training Contest 1
 
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#include <algorithm>#include <cstring>#include <cstdio>#include <queue>#include <iostream>#include <vector>#include <string>#include <map>#include <cmath>#define REP(a,b,c) for(a=b;a<=c;a++)using namespace std;typedef long long ll;const ll inf= 1e8;const int N=52;const int M=16 ;double dp[N][5];double p[3][3]={{0.5,0.375,0.125},                {0.25,0.125,0.625},                {0.25,0.375,0.375}                };double q[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 n;int hash(string str){    if(str=="Dry")return 0;    else if(str=="Dryish")return 1;    else if(str=="Damp")return 2;    else return 3;}int x[N],pa[N][3],ans[N];char * weather[3]={(char*)"Sunny",(char*)"Cloudy",(char*)"Rainy"};int run(){    string str;    int _,cas=0;    cin>>_;    while(_--)    {        cin>>n;        for(int i=1;i<=n;++i){            cin>>str;            x[i] = hash(str);        }        for(int i=1;i<=n;++i){            for(int j=0;j<3;++j)                dp[i][j]= log(0);        }        memset(pa,0,sizeof(pa));        dp[1][0] = log(0.63) + log(q[0][x[1]]);        dp[1][1] = log(0.17) + log(q[1][x[1]]);        dp[1][2] = log(0.20) + log(q[2][x[1]]);        for(int i=2;i<=n;++i)            for(int j=0;j<3;++j)                for(int k=0;k<3;++k){                    double temp=dp[i-1][k]+log(p[k][j])+log(q[j][x[i]]);                    if(temp > dp[i][j] )                    {                        dp[i][j]=temp;                        pa[i][j]=k;                    }                }        int pos=0;        for(int i=0;i<3;++i)            if(dp[n][i] > dp[n][pos]) pos=i;        ans[n]=pos;        for(int i=n-1;i>=1;--i){            pos=pa[i+1][pos];            ans[i]=pos;        }            printf("Case #%d:\n",++cas);        for(int i=1;i<=n;++i)            printf("%s\n",weather[ans[i]]);    }    return 0;}int main(){    //freopen("in.txt","r",stdin);    ios::sync_with_stdio(0);    return run();}

 

 


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