POJ 3744.Scout YYF

402


题目



点击显/隐题目

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.


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].


For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.


1 0.5
2
2 0.5
2 4


0.5000000
0.2500000




题解



不考虑地雷的情况下,可以很容易得到递推式:
dp(i) = p*dp(i-1)+(1-p)*dp(i-2)

然而由于距离可以达到100000000,直接去计算需要非常多的内存

根据概率的知识,可以知道,能够通过每一个雷区的概率等于1-踩雷的概率
而通过整个区域的概率就是通过每个雷区的概率的乘积

根据雷区的位置,可以把整个区域分成多段,每一段有且仅有一个地雷
计算每一段的概率即可

不过每一段的长度也有可能是非常长的,要解决它有两种思路:
高中数学优化高等数学优化

对于 $a_n = p times a_{n-1}+(1-p) times a_{n-2}$ ,求其通项公式
化简步骤如下:

$$ \begin{align*} a_n &= p \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n &= (1-(1-p)) \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n &= a_{n-1} - (1-p) \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n - a_{n-1} &= -(1-p) (a_{n-1} - a_{n-2}) \\ \end{align*} $$

这时,已经可以看出等号两边的结构相等
设一个辅助数列
$$ \begin{align*} 令 b_n &= a_{n+1}-a_{n} \\ \\ 有 b_n &= (p-1) \times b_{n-1} \\ 可得 b_n &= b_1 \times (p-1)^n-1 \\ b_{n-1} &= b_1 \times (p-1)^{n-2} \\ b_{n-1} &= a_{n} - a_{n-1}\\ \\ a_{n} - a_{n-1} &= (a_2 - a_1) \times (p-1)^{n-2} \\ a_{n-1} - a_{n-2} &= (a_2 - a_1) \times (p-1)^{n-3} \\ & ...\\ a_{2} - a_{1} &= (a_2 - a_1) \times (p-1)^{0} \\ \\ a_{n} - a_{1} &= \sum_{i=0}^{n-2} ((a_2 - a_1) \times (p-1)^i) \\ a_{n} - a_{1} &= (a_2 - a_1) \times \sum_{i=0}^{n-2} (p-1)^i) \\ a_{n} - a_{1} &= (a_2 - a_1) \times \frac {1 \times (1-(p-1)^{n-1})} {1-(p-1)} \\ a_{n} &= a_{1} + (a_2 - a_1) \times \frac {1-(p-1)^{n-1}} {2-p} \\ \\ 其中 a_1 &= 1 , a_2 = p\\ a_{n} &= 1 + (p-1) \times \frac {1-(p-1)^{n-1}} {2-p} \\ \end{align*} $$

也即,我们有了dp的通项公式
dp[i] = 1 + (p-1)*(1-(p-1)^(n-1)/(2-p))

平方部分使用快速幂计算即可


当然,也可以使用高等数学部分的矩阵乘法

\begin{pmatrix} p & 1-p\\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} dp[i] \\ dp[i-1] \end{pmatrix} = \begin{pmatrix} p \times dp[i] + (1-p) \times dp[i-1] \\ dp[i] \end{pmatrix} = \begin{pmatrix} dp[i+1] \\ dp[i] \end{pmatrix} ${% endraw %} 那么可以推出dp[i]的表达式为 {% raw %}$ \begin{pmatrix} dp[i]\\ dp[i-1] \end{pmatrix} = \begin{pmatrix} p & 1-p\\ 1 & 0 \end{pmatrix} ^ {n-2} \times \begin{pmatrix} p\\ 1 \end{pmatrix} ${% endraw %} 使用快速矩阵幂计算即可,需要注意两个雷紧挨着的时候的情况 # 代码 {% fold 点击显/隐递推方法代码 %}```cpp Scout YYF递推 https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份 /*/ #define debug #include //*/ #include #include #include #include #include #include using namespace std; const int maxn = 15; double pow(double a, int n) { if (n == 0) return 1.0; if (n == 1) return a; double ans = pow(a, n / 2); ans = ans * ans; if (n & 1) ans *= a; return ans; } double f(int n,double p){ return 1 + (p - 1) * (1 - pow(p - 1, n - 1)) / (2 - p); } int mines[maxn]; int main() { #ifdef debug freopen("in.txt", "r", stdin); int START = clock(); #endif cin.tie(0); cin.sync_with_stdio(false); int n; double p; while (cin >> n >> p) { mines[0] = 0; for (int i = 1; i <= n; i++) cin >> mines[i]; sort(mines + 1, mines + 1 + n); double ans = 1; for (int i = 1; i <= n; i++) { int dis = mines[i] - mines[i - 1]; double dp = f(dis,p); //cout <<"\t "< //*/ #include #include #include #include #include #include using namespace std; const int maxn = 15; class Matrix { public: static const int LINE = 2; //行列数 double matrix[LINE][LINE]; Matrix() { for (int i = 0; i < LINE; i++) for (int j = 0; j < LINE; j++) matrix[i][j] = 0; } Matrix operator*(Matrix rhs) { return mul(*this, rhs); } Matrix operator^(int n) { return pow(*this, n); } static Matrix mul(Matrix a, Matrix b) { Matrix ans; for (int i = 0; i < LINE; i++) for (int j = 0; j < LINE; j++) { ans.matrix[i][j] = 0; for (int k = 0; k < LINE; k++) ans.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j]; } return ans; } static Matrix pow(Matrix a, int n) { if (n == 0) { Matrix E; for (int i = 0; i < LINE; i++) E.matrix[i][i] = 1; return E; } if (n == 1) return a; Matrix ans = pow(a, n / 2); ans = ans * ans; if (n & 1) return ans * a; return ans; } void print() { for (int i = 0; i < LINE; i++) { printf("|"); for (int j = 0; j < LINE; j++) printf("%f ", matrix[i][j]); printf("|\n"); } printf("\n"); } }; int mines[maxn]; int main() { #ifdef debug freopen("in.txt", "r", stdin); int START = clock(); #endif cin.tie(0); cin.sync_with_stdio(false); int n; double p; while (cin >> n >> p) { mines[0] = 0; for (int i = 1; i <= n; i++) cin >> mines[i]; sort(mines + 1, mines + 1 + n); double ans = 1; for (int i = 1; i <= n; i++) { int dis = mines[i] - mines[i - 1]; if (dis == 1) { ans = 0; break; } Matrix dp, pro; dp.matrix[0][0] = p; dp.matrix[1][0] = 1; pro.matrix[0][0] = p; pro.matrix[0][1] = 1 - p; pro.matrix[1][0] = 1; dp = Matrix::pow(pro, dis - 2) * dp; dp.print(); ans *= (1 - dp.matrix[0][0]); } cout << fixed << setprecision(7) << ans << endl; } #ifdef debug printf("Time:%.3fs.\n", double(clock() - START) / CLOCKS_PER_SEC); #endif return 0; } ```
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