P6835 [Cnoi2020] 线形生物 部分分推导

P6835 [Cnoi2020] 线形生物 部分分推导（无证明）

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链接

P6835 [Cnoi2020] 线形生物。

推导

Subtask1

返祖图中所有点都有自环且所有边均为自环（未画出），总图形如：

\[f_i = \frac{d_i}{d_i+1} f_i + \frac{1}{d_i+1} f_{i+1} + 1 \\ f_i = f_{i+1} + d_i + 1 \\ \]

Subtask2

返祖图中所有点均向且仅向自己的前驱连边，特别地，\(1\) 号节点的前驱是 \(1\) 号节点，总图形如：

\[\because f_{i} = \frac{d_{i}}{d_{i}+1}f_{i-1} + \frac{1}{d_{i}+1}f_{i+1} +1,令 f_{i} = k_{i}f_{i+1} + b_{i} , k_{1} = 1 , b_{1} = d_{1} + 1 \\ f_{i} = \frac{d_{i}}{d_{i}+1}(k_{i-1}f_{i} + b_{i-1}) + \frac{1}{d_{i}+1}f_{i+1} +1 \\ f_{i} = \frac{f_{i+1} + (d_{i}b_{i-1} + d_{i} + 1)}{d_{i}+1-d_{i}k_{i-1}} \\ \therefore k_{i} = \frac{1}{d_{i}+1-d_{i}k_{i-1}} , b_{i} = \frac{d_{i}b_{i-1} + d_{i} + 1}{d_{i}+1-d_{i}k_{i-1}} \\ \]

Subtask3

返祖图中所有点均向且仅向 \(1\) 号节点连边，总图形如：

\[f_{i} = \frac{1}{d_{i} + 1}f_{i+1} + \frac{d_{i}}{d_{i} + 1} f_1 + 1 , f_{i+1} = 0 \\ f_{i-1} = \frac{1}{d_{i-1} + 1}(\frac{d_{i}}{d_{i} + 1} f_1 + 1) + \frac{d_{i-1}}{d_{i-1} + 1} f_1 + 1 \\ f_{i-2} = \frac{1}{d_{i-2} + 1}(\frac{1}{d_{i-1} + 1}(\frac{d_{i}}{d_{i} + 1} f_1 + 1) + \frac{d_{i-1}}{d_{i-1} + 1} f_1 + 1) + \frac{d_{i-2}}{d_{i-2} + 1} f_1 + 1 \\ ...... \\ f_{i-1} = \frac{(d_{i} + 1)(d_{i-1} + 1) - 1}{(d_{i} + 1)(d_{i-1} + 1)} f_1 + \frac{d_{i-1} + 2}{d_{i-1} + 1}\\ f_{i-2} = \frac{(d_{i} + 1)(d_{i-1} + 1)(d_{i-2} + 1) - 1}{(d_{i} + 1)(d_{i-1} + 1)(d_{i-2} + 1)} f_1 + \frac{(d_{i-1} + 1)(d_{i-2} + 1) + (d_{i-1} + 1) + 1}{(d_{i-1} + 1)(d_{i-2} + 1)}\\ ...... \\ f_1 = \frac{\prod_{i=1}^n(d_i+1) - 1}{\prod_{i=1}^n(d_i+1)} f_1 + \frac{\sum_{i=1}^{i-1}\prod_{j=i}^{i-1}(d_i+1) + 1}{\prod_{i=1}^{n-1}(d_i+1)}\\ \prod_{i=1}^n(d_i+1)f_1 = [\prod_{i=1}^n(d_i+1) - 1] f_1 + \sum_{i=1}^{n}\prod_{j=i}^{n}(d_i+1)\\ f_1 = \sum_{i=1}^{n}\prod_{j=i}^{n}(d_j+1)\\ \]

Subtask5

\[E_{i \to i+1} = \frac{1}{d_i+1} + \frac{1}{d_i+1}\sum_{e(i,j) \in S} (E_{j \to i+1} + 1) \\ E_{i \to i+1} = 1 + \frac{1}{d_i+1}\sum_{e(i,j) \in S} E_{j \to i+1} \\ E_{i \to i+1} = 1 + \frac{1}{d_i+1}\sum_{e(i,j) \in S} \sum_{k=j}^{i}E_{k \to k+1} \\ f_i = E_{i \to i+1} , s_i = \sum_{i=1}^{i} f_i\\ f_i = 1 + \frac{1}{d_i+1}\sum_{e(i,j) \in S} (s_i-s_{j-1}) \\ f_i = 1 + \frac{1}{d_i+1}\sum_{e(i,j) \in S} (s_{i-1}-s_{j-1}) + \frac{d_i}{d_i+1} f_i\\ f_i = d_i + 1 + \sum_{e(i,j) \in S} (s_{i-1}-s_{j-1})\\ ANS = \sum_{i=1}^{n} f_i \\ \]

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（不敢保证正确性）