POJ 3744 Scout YYF I 矩阵快速幂优化--概率dp

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本文介绍了一种使用矩阵快速幂算法解决雷区穿越问题的方法，旨在计算角色 YYF 在特定概率下成功避开雷区的概率。文章详细展示了如何通过状态转移方程转化为矩阵运算，并提供了完整的 C++ 实现代码。

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Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5416		Accepted: 1491

Description

YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1- p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

Input

The input contains many test cases ended with EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

Output

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

Sample Input

Sample Output

0.5000000
0.2500000

Source

POJ Monthly Contest - 2009.08.23, Simon

有n个雷放在1~100000000的位置上，初始位置是1，走一步的概率是p，走两步的概率是1-p，求顺利通过所有雷的概率。

状态转移方程：dp[i]=p*dp[i-1]+(1-p)*dp[i-2],如果直接算会MLE或者TLE，可以转换成用矩阵快速幂求解。

那么可以分别求出从起点到第一个雷的概率，然后求出第一个雷的位置+1到第二个雷的概率，依次分别求出不睬到雷的概率(1-踩到雷的概率），然后累乘就是答案。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
double p;
int pos[17];
struct Matrax
{
    double m[2][2];
}a,per,tmp;
void init()//建立矩阵
{
    a.m[0][0]=p;a.m[0][1]=1-p;
    a.m[1][0]=1;a.m[1][1]=0;
    per.m[0][0]=1;per.m[0][1]=0;
    per.m[1][0]=0;per.m[1][1]=1;
}
Matrax multi(Matrax a,Matrax b)//矩阵相乘
{
    Matrax c;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<2;k++)
                c.m[i][j]+=a.m[i][k]*b.m[k][j];
        }
        return c;
}
Matrax power(int k)//矩阵快速幂
{
    Matrax pp=a,ans=per;
    while(k)
    {
        if(k&1){ans=multi(ans,pp);k--;}
        else {k>>=1;pp=multi(pp,pp);}
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d%lf",&n,&p)!=EOF)
    {
        bool no=false;
        double ans=1;
        memset(pos,0,sizeof(pos));
        for(int i=1;i<=n;i++)
            scanf("%d",&pos[i]);
        sort(pos+1,pos+n+1);
        for(int i=1;i<=n;i++)
            if(pos[i]==1||pos[i]==pos[i-1]){no=true;break;}
        if(no){printf("%.7lf\n",0);continue;}

        init();
        tmp=power(pos[1]-1);//求从1到第一个雷的概率
        ans*=(1-tmp.m[0][0]);//求不到第一个雷的概率
        for(int i=2;i<=n;i++)
        {
            if(pos[i]==pos[i-1])continue;
            tmp=power(pos[i]-pos[i-1]-1);//求从第i-1个雷后面一步到第i个雷的概率
            ans*=(1-tmp.m[0][0]);//不到第i个雷的概率
        }
        printf("%.7lf\n",ans);
    }
    return 0;
}