[TOJ 3978] Probability II

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#概率

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本文探讨了一个关于Bob如何向Alice表白的概率问题。Bob可以在多个点进行表白，每个点的成功概率不同，失败则回到起点。文章通过动态规划的方法，计算了在限定表白次数下Bob达到终点期望展示次数。

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3978. Probability II

Before read this problem, you should read the problem "Probability I" to get some background information.

In fact, Bob is far more intelligent than we have imagined. He will vindicate to Alice at any point which means he may say to Alice that he love her directly. Bob knows that if he do this at the point i, he can success with the probabilityp_i4. But if he fails, he must be back to the point 1. And at the point n he don’t need Vindicate to Alice. Bob is not brave enough as ZHENGWEI, so he decides that he will vindicate at most m times. If he all failed at the m times, he won’t vindicate again which means he will then try his best to show step by step until he reaches the point n. Now we want to know how many expect times Bob have to show using his best strategy.(When Bob directly say to Alice, it is also once show).

Input

There will be several test cases, and each test case input as follows:
The first line input n(1≤n≤1000), m(1≤m≤1000). Then there will be n-1 lines following. Each of the following lines contains four real number p_i1, p_i2, p_i3, p_i4. You can trust that p_i1+p_i2+p_i3=1(p_i1, p_i2 and p_i3 have been described in the problem "Probability I". 0 ≤ p_i1, p_i2, p_i3, p_i4≤ 1, p_i3=0 at point n-1).

Output

For each test case, please output the expect times that Bob has to show to Alice. Answers must be rounded to one digits after the decimal point. If he can’t reach at the point n, please output "impossible".

Sample Input

2 1

0.3 0.7 0.0 1.0

3 1

0.2 0.3 0.5 0.2

0.3 0.7 0.0 1.0

Sample Output

1.0

1.6

dp再加上一些概率上的问题先推公式，然后每次可以有两种选择，从后向前递推求解。

#include<iostream>
#include<cstring>
#include<iomanip>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
int n,m;
const double inf=101010101010.0;
double dp[1200][1200];
double p[1200][4];
int main()
{
    while(cin>>n>>m)
    {
        memset(dp,0,sizeof(dp));
        bool flag=1;
        for(int i=1;i<=n-1;i++)
            cin>>p[i][0]>>p[i][1]>>p[i][2]>>p[i][3];
        for(int j=m;j>=0;j--)
        for(int i=n-1;i>=1;i--)
                if(j==m)
                    dp[i][j]=(dp[i+1][j]*p[i][1]+dp[i+2][j]*p[i][2]+1)/(1-p[i][0]);
                else
                    dp[i][j]=min((dp[i+1][j]*p[i][1]+dp[i+2][j]*p[i][2]+1)/(1-p[i][0]),dp[1][j+1]*(1-p[i][3])+1);
        double ans=inf;
        for(int i=0;i<=m;i++)
            ans=min(ans,dp[1][i]);
        if(ans>101010100)
            puts("impossible");
        else
            printf("%.1f\n",ans);
    }
}