hdu 4865

来源：互联网发布：python余弦相似度编辑：程序博客网时间：2024/05/29 09:23

当时比赛时没看这道题，后来看了下，感觉完全可以当做是条件概率之类的事情来做？其实这个是典型的隐马尔科夫模型的应用，这篇文章介绍的很不错http://blog.csdn.net/likelet/article/details/7056068

根本思想就是到第i天最优路径可以用第i-1天的最优路径推出来，也就是所谓的无后效性，其本质类似于递推，结合下概率方面的知识，递推一下就可以了

代码https://github.com/mlz000/hdu/blob/master/4865(Markov).cpp

#include<iostream>//Hidden Markov Model#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<map>using namespace std;double p1[3][3]={    {0.5 ,0.375,0.125},    {0.25,0.125,0.625},    {0.25,0.375,0.375}};double p2[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 f[50][5];int pre[50][5];map<string,int>mp;char ans[3][10]={"Sunny","Cloudy","Rainy"};void print(int i,int j){if(i==0)return;print(i-1,pre[i][j]);printf("%s\n",ans[j]);}int main(){mp["Dry"]=0,mp["Dryish"]=1,mp["Damp"]=2,mp["Soggy"]=3;int T,n;string s;cin>>T;for(int tt=1;tt<=T;++tt){f[1][0]=0.63,f[1][1]=0.17,f[1][2]=0.20;printf("Case #%d:\n",tt);cin>>n;for(int i=1;i<=n;++i){cin>>s;int k=mp[s];for(int j=0;j<3;++j){if(i==1)f[i][j]=log(f[i][j])+log(p2[j][k]);else{f[i][j]=-1000000.0;for(int l=0;l<3;++l){if(f[i][j]<f[i-1][l]+log(p1[l][j])+log(p2[j][k])){f[i][j]=f[i-1][l]+log(p1[l][j])+log(p2[j][k]);pre[i][j]=l;}}}}}int last=0;if(f[n][1]>f[n][last])last=1;if(f[n][2]>f[n][last])last=2;print(n,last);}return 0;}

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