POJ 3744 Scout YYF(概率DP+矩陣快速冪）

題目鏈接：點擊打開鏈接

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

題意：一條路有n個陷阱，起始點在1，每次走一步的概率是p，兩步的概率是1-p，給出n個陷阱的位置，在[1, 100000000]之間，求安全通過的概率。

思路：設點dp[n]爲到達n的概率，則dp[n]=p*dp[n-1]+(1-p)*dp[n-2]

          但是由於座標範圍很大，直接這樣求是不行的，而且當中的某些點還存在地雷。輸入地雷座標x[i]

          分段求通過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終的答案。

        求dp[i]的時候，需要用到矩陣快速冪。那麼通過每一段的概率就是1-dp[x[i]]，同時需要保證每一段路的起點的概率是1,如dp[1]=1；

代碼：

#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);
    }
}