Count BFS Graph

Count BFS Graph

题目信息

Luogu CF1906J、Codeforces 1906J

题面翻译

对于一个 \(n\) 个节点的无向图的邻接矩阵 \(M\)，满足 \(M_{i,i}=0,M_{u,v}=M_{v,u}\)，\(M_{i,j}=1\) 表示 \(M_{i,j}\) 右边，进行下面的 bfs 生成 \(A\) 数组。

BFS():
	清空数组 A, 清空队列 Q
        在 A 中加入 1, 在 Q 中加入 1
        while Q 不为空:
            令 u 为 Q 的队头, 并弹出对头
            将 v 从 1 遍历到 n:
            	if(M[u][v] == 1 并且 v 不在 A 中):
                	Q 中加入 v，在 A 末尾加入 v

给定 \(A\) 数组，求多少个 \(M\) 满足条件。

\(1\le n\le5000\)，\(A\) 为 \(1\sim n\) 的排列。

答案对 \(998\,244\,353\) 取模。

题目描述

You are currently researching a graph traversal algorithm called the Breadth First Search (BFS). Suppose you have an input graph of $ N $ nodes (numbered from $ 1 $ to $ N $ ). The graph is represented by an adjacency matrix $ M $ , for which node $ u $ can traverse to node $ v $ if $ M_{u, v} $ is $ 1 $ , otherwise it is $ 0 $ . Your algorithm will output the order the nodes are visited in the BFS. The pseudocode of the algorithm is presented as follows.

BFS(M[1..N][1..N]):
        let A be an empty array
        let Q be an empty queue

        append 1 to A
        push 1 to Q

        while Q is not empty:
            pop the front element of Q into u
            for v = 1 to N:
                if M[u][v] == 1 and v is not in A:
                    append v to A
                    push v to Q

        return A

During your research, you are interested in the following problem. Given an array $ A $ such that $ A $ is a permutation of $ 1 $ to $ N $ and $ A_1 = 1 $ . How many simple undirected graph with $ N $ nodes and adjacency matrix $ M $ such that $ \text{BFS}(M) = A $ ? Since the answer can be very large, calculate the answer modulo $ 998,244,353 $ .

A simple graph has no self-loop ( $ M_{i, i} = 0 $ for $ 1 \leq i \leq N $ ) and there is at most one edge that connects a pair of nodes. In an undirected graph, if node $ u $ is adjacent to node $ v $ , then node $ v $ is also adjacent to node $ u $ ; formally, $ M_{u, v} = M_{v, u} $ for $ 1 \leq u < v \leq N $ .

Two graphs are considered different if there is an edge that exists in one graph but not the other. In other words, two graphs are considered different if their adjacency matrices are different.

输入格式

The first line consists of an integer $ N $ ( $ 2 \leq N \leq 5000 $ ).

The second line consists of $ N $ integers $ A_i $ . The array $ A $ is a permutation of $ 1 $ to $ N $ and $ A_1 = 1 $ .

输出格式

Output an integer representing the number of simple undirected graphs with $ N $ nodes and adjacency matrix $ M $ such that $ \text{BFS}(M) = A $ . Since the answer can be very large, output the answer modulo $ 998,244,353 $ .

样例 #1

样例输入 #1

3
1 2 3

样例输出 #1

3

样例 #2

样例输入 #2

3
1 3 2

样例输出 #2

1

样例 #3

样例输入 #3

5
1 3 2 4 5

样例输出 #3

17

样例 #4

样例输入 #4

11
1 2 3 4 5 6 7 8 9 10 11

样例输出 #4

379394847

提示

Explanation for the sample input/output #1

The following illustration shows all graphs that satisfy the requirements.

Explanation for the sample input/output #2

The only graph that satisfies the requirements is a graph with two edges: one that connects nodes $ 1 $ and $ 3 $ , and another one that connects nodes $ 3 $ and $ 2 $ .

思路分析

这道题结构性很强。所以我们可以考虑 DP。
对于每一个数，其实我们只需要看它能连几个数就行了。
定义：\(dp_{i,j}\) 表示以 \(A_i\) 是队首，\(A_j\) 是队尾的方案数。

决策粒度变小。

\(dp_{i,j,0/1}\)

tag

Codeforces、题解
数学、容斥原理