【POJ3744】Scout YYF I

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#矩阵快速幂 #概率DP

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本文介绍了一种算法，用于计算角色 YYF 在给定概率下安全通过布满地雷的道路的概率。通过分段处理和矩阵快速幂的方法，解决了路径中存在多个地雷的问题。

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Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10629		Accepted: 3139

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

解析：

令f[i]表示在没有地雷的情况下，走i距离的几率，可得f[i]=f[i-1]*p+f[i-2]*(1-p)，但是当对于有地雷的位置需要特判且会T，我们可以分段处理。按照地雷分段1−x[1],x[1]+1−x[2],……1−x[1],x[1]+1−x[2],……,分段矩阵快速幂即可，求出每一段通过的概率是f[x[i]+1]=1−f[x[i]]f[x[i]+1]=1−f[x[i]]，乘起来就行了。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
using namespace std;

int n,pos[13];
double p,ans=1;

inline void mul(double ans[2],double a[2][2])
{
	double c[2];
	memset(c,0,sizeof(c));
	for(int i=0;i<2;i++)
	  for(int k=0;k<2;k++)
	    c[i]+=ans[k]*a[k][i];
	memcpy(ans,c,sizeof(c));
}

inline void mulself(double a[2][2])
{
	double c[2][2];
	memset(c,0,sizeof(c));
	for(int i=0;i<2;i++)
	  for(int j=0;j<2;j++)
	    for(int k=0;k<2;k++)
	      c[i][j]+=a[i][k]*a[k][j];
	memcpy(a,c,sizeof(c));
}

inline double solve(int l,int r)
{
	double ans[2]={p,1},a[2][2]={{p,1.0},{1-p,0}};
	int n=r-l-1;
	if(l==r) return ans[1];
	while(n)
	{
	  if(n&1) mul(ans,a);
	  n>>=1;
	  mulself(a);
	}
	return ans[0];
}

int main()
{
	while(~scanf("%d%lf",&n,&p))
	{
	  ans=1;
	  for(int i=1;i<=n;i++) scanf("%d",&pos[i]);
	  sort(pos+1,pos+n+1);
	  n=unique(pos+1,pos+n+1)-pos-1;
	  ans*=(1.0-solve(1,pos[1]));
	  for(int i=2;i<=n;i++) ans*=(1.0-solve(pos[i-1]+1,pos[i]));
	  printf("%.7f\n",ans);
	}
	return 0;
}