[Poj3744]Scout YYF I （概率dp + 矩阵乘法）

Scout YYF I

---

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9552		Accepted: 2793

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

---

 0.5

 0.5
 

Sample Output

---

0.5000000
0.2500000

分析：

概率dp是需要顺着推的，定义dp[i]表示到i的概率。当i有雷时，dp[i] = 0；

否则 dp[i] = dp[i - 1] * p + dp[i - 1] * （1 - p）

最后输出dp[dis[n] + 1]即行。

因为n很大，转移又固定，可以联想到矩乘 + 快速幂。于是就0MS过了

坑： 输入雷不一定有序，需要排序，精度输出需要%.7f  不能 %.7lf。

AC代码：

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
using namespace std;
int n;
int dis[];
struct fi{
    double data[][];
    fi(){
        for(int i = ;i < ;i++){
            for(int j = ;j < ;j++){
                data[i][j] = ;
            }
        }
    }
}A,T,O;
inline fi operator *(fi A,fi B){
    fi t;
    for(int i = ;i < ;i++){
        for(int j = ;j < ;j++){
            for(int k = ;k < ;k++){
                t.data[i][j] += A.data[i][k] * B.data[k][j];
            }
        }
    }
    return t;
}
fi cmd(fi C,int k){
    fi D = O;
    while(k){
        if(k & )D = D * C;
        k >>= ;
        C = C * C;
    }
    return D;
}
double p;
int main(){
    for(int i = ;i < ;i++)O.data[i][i] = 1.0;
    while(~scanf("%d %lf",&n,&p)){
    T.data[][] =  (1.0 - p);
    T.data[][] = p;
    T.data[][] = 1.0;
    T.data[][] = 0.0;
        for(int i = ;i < ;i++){
            for(int j = ;j < ;j++){
                A.data[i][j] = 0.0;
            }
        }
        A.data[][] = 1.0;
    for(int i = ;i <= n;i++){
        scanf("%d",&dis[i]);
    }
    sort(dis + ,dis + n + );
    for(int i = ;i <= n;i++){
        A = A * cmd(T,dis[i] - dis[i - ] - );
        A.data[][] = ;
        A = A * T;
    }
    printf("%.7f\n",A.data[][]);
  }
}