Score : $900$ points
Given are simple undirected graphs $X$, $Y$, $Z$, with $N$ vertices each and $M_1$, $M_2$, $M_3$ edges, respectively. The vertices in $X$, $Y$, $Z$ are respectively called $x_1, x_2, \dots, x_N$, $y_1, y_2, \dots, y_N$, $z_1, z_2, \dots, z_N$. The edges in $X$, $Y$, $Z$ are respectively $(x_{a_i}, x_{b_i})$, $(y_{c_i}, y_{d_i})$, $(z_{e_i}, z_{f_i})$.
Based on $X$, $Y$, $Z$, we will build another undirected graph $W$ with $N^3$ vertices. There are $N^3$ ways to choose a vertex from each of the graphs $X$, $Y$, $Z$. Each of these choices corresponds to the vertices in $W$ one-to-one. Let $(x_i, y_j, z_k)$ denote the vertex in $W$ corresponding to the choice of $x_i, y_j, z_k$.
We will span edges in $W$ as follows:
Then, let the weight of the vertex $(x_i, y_j, z_k)$ in $W$ be $1,000,000,000,000,000,000^{(i +j + k)} = 10^{18(i + j + k)}$. Find the maximum possible total weight of the vertices in an independent set in $W$, and print that total weight modulo $998,244,353$.
Input is given from Standard Input in the following format:
$N$ $M_1$ $a_1$ $b_1$ $a_2$ $b_2$ $\vdots$ $a_{M_1}$ $b_{M_1}$ $M_2$ $c_1$ $d_1$ $c_2$ $d_2$ $\vdots$ $c_{M_2}$ $d_{M_2}$ $M_3$ $e_1$ $f_1$ $e_2$ $f_2$ $\vdots$ $e_{M_3}$ $f_{M_3}$
Print the maximum possible total weight of an independent set in $W$, modulo $998,244,353$.
2 1 1 2 1 1 2 1 1 2
46494701
The maximum possible total weight of an independent set is that of the set $(x_2, y_1, z_1)$, $(x_1, y_2, z_1)$, $(x_1, y_1, z_2)$, $(x_2, y_2, z_2)$. The output should be $(3 \times 10^{72} + 10^{108})$ modulo $998,244,353$, which is $46,494,701$.
3 3 1 3 1 2 3 2 2 2 1 2 3 1 2 1
883188316
100000 1 1 2 1 99999 100000 1 1 100000
318525248