poj 3744 Scout YYF I（矩阵优化概率DP）

Scout YYF I

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

1 0.5
2
2 0.5
2 4

Sample Output

0.5000000
0.2500000

给n个有地雷的位置，给出走1步的概率和2步的概率，求能顺利通过这条路的概率。

首先对这n个点相邻两点排序分段[1,a[0]],[a[0]+1,a[1]] .......，假设k点有地雷，那么这个人肯定走k+1点，那么把n个点分成n段，每段起点的概率都相同，最后把每段不走地雷点的概率相乘就是结果。

由于区间长度特别大，不能Ｏ(n)递推,要用矩阵快速幂优化，很容易推出递推公式f(n)=f(n-1)*p+f(n-2)*(1-p),n点只能从n-1点和n-2点到。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct matrix
{
    double ma[4][4];
};
matrix multi(matrix x,matrix y)
{
    matrix ans;
    memset(ans.ma,0,sizeof(ans.ma));
    for(int i=1; i<=2; i++)
    {
        for(int j=1; j<=2; j++)
        {
            if(x.ma[i][j])
            {
                for(int k=1; k<=2; k++)
                    ans.ma[i][k]=ans.ma[i][k]+x.ma[i][j]*y.ma[j][k];
            }
        }
    }
    return ans;
}
matrix pow(matrix a,int m)
{
    matrix ans;
    for(int i=1; i<=2; i++)
    {
        for(int j=1; j<=2; j++)
        {
            if(i==j)
                ans.ma[i][j]=1;
            else
                ans.ma[i][j]=0;
        }
    }
    while(m)
    {
        if(m&1)
            ans=multi(ans,a);
        a=multi(a,a);
        m=m>>1;
    }
    return ans;
}
int main()
{
    int a[15];
    int n;
    double p;
    while(~scanf("%d%lf",&n,&p))
    {
        double ans=1;
        matrix b,d;
        matrix c;
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        b.ma[1][1]=p;
        b.ma[1][2]=1-p;
        b.ma[2][1]=1;
        b.ma[2][2]=0;
        memset(c.ma,0,sizeof(c.ma));
        c.ma[1][1]=p;
        c.ma[2][1]=1;
        if(a[0]==1||a[0]==2)
        {
            if(a[0]==1)
                ans=0;
            else
                ans=ans*(1-p);
        }
        else
        {
            d=pow(b,a[0]-2);
            d=multi(d,c);
            ans=ans*(1-d.ma[1][1]);
        }
        for(int i=1; i<n; i++)
        {
            int temp=a[i]-a[i-1];
            if(temp==1||temp==2)
            {
                if(temp==1)
                    ans=0;
                else
                    ans=ans*(1-p);
            }
            else
            {
                d=pow(b,temp-2);
                d=multi(d,c);
                ans=ans*(1-d.ma[1][1]);
            }
        }
        printf("%.7f\n",ans);
    }
    return 0;
}