POJ 3744 Scout YYF I （概率dp+矩阵优化）

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FFmpeg魔改指南：让RTMP协议完美兼容H265的三种方法（含SRS适配技巧）本文详细介绍了三种让RTMP协议兼容H265编码的FFmpeg魔改方案，包括基于官方补丁的标准改造、Intel的SVT-HEVC集成方案以及金山云的自定义扩展方案。文章重点解析了各方案的实现原理与适配技巧，并提供了在SRS 6.0服务器上进行深度配置和性能调优的完整实战指南，帮助开发者解决H265推流至传统RTMP服务器的兼容性问题。阅读详情

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Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5504		Accepted: 1522

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

题意：当前点是 1 ，有n个放置的雷，求不踩雷的概率通过

思路: 参照 kuangbin 大神博客 http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710586.html

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 15

int a[N];
int n;
double p;

struct mat{
   double A[3][3];
   mat(){ mem(A,0); }

   mat operator * (mat b)
   {
      mat c;
      int i,j,k;

   	  fre(i,0,3)
   	   fre(j,0,3)
   	   {
            fre(k,0,3)
   	   	     c.A[i][j]+=A[i][k]*b.A[k][j];
   	   }

   	   return c;
   }

}s;

mat pow_mul(mat s,int n)
{
     mat ans;
     int i,j;

     fre(i,0,3)
     	ans.A[i][i]=1;

    while(n)
	{
		if(n&1) ans=ans*s;
		s=s*s;
		n>>=1;
	}

	return ans;
}

int main()
{
	int i,j;

	while(~scanf("%d%lf",&n,&p))
	{
		fre(i,0,n)
		 sf(a[i]);

		 double ans=1;

		 s.A[0][0]=p;
		 s.A[0][1]=1-p;
		 s.A[1][0]=1;
		 s.A[1][1]=0;

		 mat temp;

		 sort(a,a+n);

		 temp=pow_mul(s,a[0]-1);   //计算走到这一个点的概率

		 ans*=(1-temp.A[0][0]);

		 fre(i,1,n)
		 {
		 	 if(a[i]==a[i-1]) continue;
		 	 temp=pow_mul(s,a[i]-a[i-1]-1);
		 	 ans*=(1-temp.A[0][0]);
		 }

       pf("%.7f\n",ans);
	}
    return 0;
}