本文主要是介绍hdu4865(概率DP)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题意：已知前一天和今天的天气概率，某天的天气概率和叶子的潮湿程度的概率，n天叶子的湿度，求n天最有可能的天气情况。思路：概率DP，dp[i][j]表示第i天天气为j的概率，状态转移如下：dp[i][j] = max(dp[i][j, dp[i-1][k]*table2[k][j]*table1[j][col] )
代码如下：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <stack>
#include <iostream>
#define N 55
using namespace std;
double dp[N][5];
char s[10];
int pre[N][5];
double table1[3][4] = { {0.6,0.2,0.15,0.05}, { 0.25,0.3,0.2,0.25 }, { 0.05,0.1,0.35,0.5 } };
double table2[3][3] = { {0.5,0.375,0.125}, {0.25,0.125,0.625}, {0.25,0.375,0.375} };
int get_col(char *s)
{if(strcmp(s,"Dry") == 0) return 0;if(strcmp(s,"Dryish") == 0) return 1;if(strcmp(s,"Damp") == 0) return 2;return 3;
}
map<int,string> mp;
int main()
{mp[0] = "Sunny";mp[1] = "Cloudy";mp[2] = "Rainy";int t,cas = 1;scanf("%d",&t);while(t--){int n;scanf("%d",&n);memset(dp,0,sizeof(dp));int i,j,k;scanf("%s",s);int col = get_col(s);dp[1][0] = 0.63*table1[0][col];dp[1][1] = 0.17*table1[1][col];dp[1][2] = 0.20*table1[2][col];for(i = 2; i <= n; i++){scanf("%s",s);col = get_col(s);for(j = 0; j < 3; j++)for(k = 0; k < 3; k++){double temp = dp[i-1][k]*table2[k][j]*table1[j][col];if(dp[i][j] < temp){dp[i][j] = temp;pre[i][j] = k;}}}double temp = 0;for(i = 0; i < 3; i++)if(dp[n][i] > temp){k = i;temp = dp[n][i];}// cout<<k<<endl;//cout<<temp<<endl;//for(i = 1; i <= n; i++){// for(j = 0; j < 3; j++)// cout<<dp[i][j]<<" ";// cout<<endl;}// for(i = 1; i <= n; i++){// for(j = 0; j < 3; j++)// cout<<pre[i][j]<<" ";
// / cout<<endl;}printf("Case #%d:\n",cas++);stack<string> q;while(n){// printf("%s\n",mp[k]);// cout<<mp[k]<<endl;q.push(mp[k]);k = pre[n][k];// cout<<k<<endl;n--;}while(!q.empty()) {cout<<q.top()<<endl;q.pop();}}return 0;
}
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