poj 3744 Scout YYF 1 (概率DP+矩阵快速幂)

F - Scout YYF I

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

题意：

有一段路，路上有n个陷阱，每一次只能向前走一步或者两步，求安全走过这段路的改路

 

分析：

设dp[i]表示安全走过第i个陷阱的概率

那么dp[i+1]=dp[i]*(1-p(走到第i+1个陷阱))

因为每次只能走一步或者两步，所有安全走过第i个陷阱后的位置一定在a[i]+1;

其中a[i]表示第i个陷阱的位置

求从a[i]+1,走到a[i+1]的概率的时候我们需要用到矩阵来经行优化

ans[i]表示走到位置i的概率

ans[i] = p*ans[i-1]+(1-p)*ans[i-2];

ans[0]=1;

都说G++比C++要好，但是本题最好用C++提交。

如果用G++提交，浮点型输出一定要用%f，否则会WA。

这是因为G++标准的浮点型输出用%f，而不是%lf。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <string.h>
using namespace std;
const int maxn=15;
int cnt[maxn];
double dp[maxn];
double p;
int n;
struct matrix
{
    double data[2][2];
};//定义成封装结构体 二维数组无法return
matrix I={1,0,0,1};
matrix multi(matrix a,matrix b)
{
    matrix c;
    memset(c.data,0,sizeof(c.data));
    for(int i=0;i<2;i++)
    for(int j=0;j<2;j++)
    for(int k=0;k<2;k++)
    c.data[i][j]+=a.data[i][k]*b.data[k][j];
    return c;
}
double pow1(matrix a,int b)
{
    matrix ans=I;
    while(b)
    {
        if(b&1)
        ans=multi(ans,a);
        b>>=1;
        a=multi(a,a);
    }
    return ans.data[0][0];
}
int main()
{
    while(cin>>n>>p)
    {
        memset(dp,0,sizeof(dp)); //该初始化的地方一定要初始化
        cnt[0]=0;  //为第一个雷做准备的
        for(int i=1;i<=n;i++)
            cin>>cnt[i];
        sort(cnt,cnt+n+1);
        dp[0]=1.0; //不是0
        matrix a={p,1-p,1,0}; //初始化矩阵，注意这个的推导过程
        for(int i=1;i<=n;i++)
        //要1减去这个概率 因为这个概率是踩上雷的
        dp[i]=dp[i-1]*(1-pow1(a,cnt[i]-cnt[i-1]-1));  //注意不要丢括号
        //注意是乘法不是加法 把每段概率乘起来
        //注意是-1不是+1
        //从上一个雷的下一个起 要走cnt[i]-cnt[i-1]-1步 才能到达下一个雷
        //雷在1和5 从2走到5 需要3步 5-1-1=3
        printf("%.7f\n",dp[n]);
    }
    return 0;
}