POJ 3744 概率dp+矩阵快速幂

Language:
Scout YYF I
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9685 Accepted: 2826
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

分析：不难看出，f[i]=p* f[i-1]+(1-p)*f[i-2],这个地推加上1e9的范围立马应该想到矩阵快速幂，于是问题就简单化了。但我们可以发现有些位置有地雷是不合法的，于是考虑分段，把地雷分成1~mines1，mines1~mines2……这样分别可以算出每一段的概率，而最后通过一段的概率就等于到达1-xi，xi等于到达xi的概率。

# include <iostream>
# include <cstdio>
# include <cmath>
# include <list>
# include <cstring>
# include <map>
# include <ctime>
# include <algorithm>
# include <queue>
using namespace std;
typedef long long ll;
int read(){
    int f=1,i=0;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {i=(i<<3)+(i<<1)+ch-'0';ch=getchar();}
    return f*i;
}
double p,a[5][5],b[5][5],res[5][5],tmp[5][5],Ans;
int mines[15],n;
inline void MUL(double a[][5],double b[][5])
{
    memset(tmp,0,sizeof(tmp));
    for(int i=0;i<=1;++i)
        for(int j=0;j<=1;++j)
            for(int k=0;k<=1;++k)
                tmp[i][j]=(tmp[i][j]+a[i][k]*b[k][j]);
    for(int i=0;i<=1;++i)
        for(int j=0;j<=1;++j) a[i][j]=tmp[i][j];
}
inline void ksm(int t)
{
    memset(res,0,sizeof(res));
    for(int i=0;i<=1;++i) res[i][i]=1;
    memcpy(a,b,sizeof(a));
    while(t)
    {
        if(t&1) MUL(res,a);
        t>>=1;
        MUL(a,a);
    }
    Ans*=(1-res[0][0]);
}
int main()
{
    freopen("lx.in","r",stdin);
    while(scanf("%d%lf",&n,&p)!=EOF)
    {
        for(int i=1;i<=n;++i) mines[i]=read();
        sort(mines+1,mines+n+1);
        a[1][1]=0,a[0][1]=1-p,a[1][0]=1,a[0][0]=p;
        b[1][1]=0,b[0][1]=1-p,b[1][0]=1,b[0][0]=p;
        Ans=1;ksm(mines[1]-1);
        for(int i=2;i<=n;++i)
        {
            if(mines[i]==mines[i-1]) continue;
            ksm(mines[i]-mines[i-1]-1);
        }
        printf("%0.7f\n",Ans);
    }
}