POJ 3744 Scout YYF(概率DP+矩阵快速幂）

题目链接：点击打开链接

Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6890		Accepted: 2004

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

Source

POJ Monthly Contest - 2009.08.23, Simon

题意：一条路有n个陷阱，起始点在1，每次走一步的概率是p，两步的概率是1-p，给出n个陷阱的位置，在[1, 100000000]之间，求安全通过的概率。

思路：设点dp[n]为到达n的概率，则dp[n]=p*dp[n-1]+(1-p)*dp[n-2]

          但是由于座标范围很大，直接这样求是不行的，而且当中的某些点还存在地雷。输入地雷座标x[i]

          分段求通过1~x[0],x[0]+1~x[1].....，x[n-2]+1~x[n-1]的概率,乘法原理相乘就是最终的答案。

        求dp[i]的时候，需要用到矩阵快速幂。那么通过每一段的概率就是1-dp[x[i]]，同时需要保证每一段路的起点的概率是1,如dp[1]=1；

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Mat
{
    double mat[2][2];
};
Mat operator *(Mat a,Mat b)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    int i,j,k;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    for(k=0;k<2;k++)
    c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
    return c;
}
Mat operator ^(Mat a,int k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    int i,j;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    if(i==j) c.mat[i][j]=1;
    while(k)
    {
        if(k&1) c=operator *(c,a);
        a=operator *(a,a);
        k>>=1;
    }
    return c;
}
int main()
{
    int n,i,x[15];double p,ans;
    Mat a,b;
    while(scanf("%d%lf",&n,&p)!=EOF)
    {
      memset(a.mat,0,sizeof(a.mat));
      a.mat[0][0]=p;a.mat[0][1]=1-p;
      a.mat[1][0]=1;a.mat[1][1]=0;
      for(i=0;i<n;i++)
      scanf("%d",&x[i]);
      sort(x,x+n);
      ans=1;
      b=operator ^(a,x[0]-1);
      ans*=(1-b.mat[0][0]);
      for(i=1;i<n;i++)
      {
         if(x[i]==x[i-1]) continue;
         b=operator ^(a,x[i]-x[i-1]-1);
         ans*=(1-b.mat[0][0]);
      }
      printf("%.7lf\n",ans);
    }
}