poj3744之矩陣快速冪+概率DP

Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4410		Accepted: 1151

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

/*分析:對於n個地雷s[1],s[2],s[3],s[4]...s[n]
假設s都是不遞減的。
假設dp[i]表示從1到達i這個位置的概率
則:
dp[s[1]-1]爲1~s[1]-1的概率//s[1]不能到達
dp[s[2]-1]爲1~s[2]-1也是1->s[1]-1->s[1]+1->s[2]-1的概率
由於最多隻能跳兩格
所以dp[s[i]+1]一定是從dp[s[i]-1]到達
然後從dp[s[i]+1]到達dp[s[i+1]-1];//這部分就可以用矩陣快速冪
另外根據公式dp[i]=p*dp[i-1]+(1-p)*dp[i-2]也可知從s[i]+1 => s[i+1]-1用矩陣快速冪求
構造初始矩陣:
p 1-p   *  dp[i]      =  dp[i+1]
1 0        dp[i-1]        dp[i]
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=10+10;
const int N=2;
int n,s[MAX];
double array[N][N],sum[N][N],p;

void InitMatrix(){
	array[0][0]=p;
	array[0][1]=1-p;
	array[1][0]=1;
	array[1][1]=0;
	for(int i=0;i<N;++i){
		for(int j=0;j<N;++j)sum[i][j]=(i == j);
	}
}

void MatrixMult(double a[N][N],double b[N][N]){
	double c[N][N]={0};
	for(int i=0;i<N;++i){
		for(int j=0;j<N;++j){
			for(int k=0;k<N;++k){
				c[i][j]+=a[i][k]*b[k][j];
			}
		}
	}
	for(int i=0;i<N;++i)for(int j=0;j<N;++j)a[i][j]=c[i][j];
}

double Matrix(int k){
	if(k<0)return 0;//表示s[i-1]~s[i]之間無位置 
	InitMatrix();//初始化矩陣
	while(k){//有k+1個位置,到達第k+1個位置所以是k次 
		if(k&1)MatrixMult(sum,array);
		MatrixMult(array,array);
		k>>=1; 
	}
	return sum[0][0];//sum[0][0]*dp[1]+sum[0][1]*dp[0]
}

int main(){
	while(~scanf("%d%lf",&n,&p)){
		for(int i=1;i<=n;++i)scanf("%d",&s[i]);
		sort(s+1,s+1+n);
		double ans=Matrix(s[1]-2);//1~s[1]-1的概率 
		for(int i=2;i<=n;++i){
			if(s[i] == s[i-1])continue;
			double temp=Matrix(s[i]-s[i-1]-2);//s[i-1]~s[i]之間有s[i]-s[i-1]-1個位置,需要走s[i]-s[i-1]-2次到達最後一個位置
			ans=ans*(1-p)*temp;//從s[i-1]-1的位置跳兩格到s[i-1]+1再到s[i]-1 
		}
		printf("%.7f\n",ans*(1-p));//在s[n]-1位置還需要跳兩格才安全了 
	}
	return 0;
}