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### Problem Statement You are given a simple conn...

创建于：2025年8月16日

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问题

Problem Statement

You are given a simple connected undirected graph with $N$ vertices numbered $1$ to $N$ and $M$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$.

For each $k=1,2,\ldots,N-1$, find the number of simple paths from vertex $1$ to vertex $N$ that contain exactly $k$ edges.

Constraints

  • $2\leq N\leq 2\times 10^5$
  • $N-1\leq M\leq N+20$
  • $1\leq u_i\lt v_i\leq N$
  • The given graph is a simple connected undirected graph.
  • All input values are integers.

Input

The input is given from Standard Input in the following format:

text
\$N\$ \$M\$
\$u\_1\$ \$v\_1\$
\$u\_2\$ \$v\_2\$
\$\\vdots\$
\$u\_M\$ \$v\_M\$

Output

Output the answers in the following format:

\$\\mathrm{ans}\_1\$ \$\\mathrm{ans}\_2\$ \$\\ldots\$ \$\\mathrm{ans}\_{N-1}\$

$\mathrm{ans}_i$ is the number of simple paths from vertex $1$ to vertex $N$ that contain exactly $i$ edges.

Sample Input 1

text
5 6
1 2
1 3
2 4
3 4
3 5
4 5

Sample Output 1

0 1 2 1

For each $k=1,2,3,4$, the simple paths from vertex $1$ to vertex $5$ that contain exactly $k$ edges are as follows.

  • $k=1$ : None
  • $k=2$ : $1\to 3\to 5$
  • $k=3$ : $1\to 2\to 4\to 5$ and $1\to 3\to 4\to 5$
  • $k=4$ : $1\to 2\to 4\to 3\to 5$

Sample Input 2

text
11 15
1 2
1 3
2 3
3 4
3 5
4 5
5 6
5 7
6 7
7 8
7 9
8 9
9 10
9 11
10 11

Sample Output 2

0 0 0 0 1 5 10 10 5 1

Sample Input 3

text
7 18
6 7
4 5
1 7
2 7
1 4
2 5
4 6
2 3
5 6
5 7
1 5
2 4
2 6
1 2
1 3
3 4
1 6
3 5

Sample Output 3

1 3 11 29 50 42

中文c++回答。

思考

Evaluating path counting methods

I need to figure out how to count simple paths from 1 to N of length k in a sparse graph (up to 2e5 nodes, 21 extra edges). Since counting simple paths is #P-hard, I could use inclusion-exclusion on cycle edges or a dynamic programming approach involving the "block-cut tree," adjusting for edge parity constraints.

Exploring path counting strategies

I’m thinking about another method: take a spanning tree and represent every s-t simple path as a tree path combined with a set of fundamental cycles using symmetric difference. The "Kirchhoff polynomial" might help here, and the cycle count (g) is small enough for dynamic programming. However, I still need to account for path lengths.

Considering graph decomposition methods

With g ≤ 20, I could try a meet-in-the-middle approach, using a BFS tree and converting to a directed acyclic graph (DAG) by orienting edges from s to t. Alternatively, I might use biconnected component decomposition and compute simple path counts by length for each block in the block-cut tree. Each path between blocks remains uniquely determined.

Refining path counting in biconnected graph

Each biconnected block has a small cyclomatic number, but the overall cyclomatic number sums to g across blocks. Now, I need to calculate the number of simple paths between two vertices within a block. I could use dynamic programming (DP) over non-tree edges or consider the "Ear decomposition" method. With M ≤ N+20, the number of cycles is limited to 21.

Optimizing graph compression and path counting

I’m considering compressing degree-2 chains into single edges to reduce complexity, which can shrink the graph size to O(g). After compression, I’ll count simple s-t paths in the compressed multigraph by using DFS with bitmasking. The number of paths could be large but manageable with optimization, and I’ll need careful handling of edges to avoid duplication in the compressed graph. This all looks promising for calculating counts efficiently.

Optimizing edge traversal and path counting

To handle edge traversal efficiently, I’ll mark original edges as visited and maintain a set of key nodes. For edge construction, I’ll avoid revisiting edges by marking them as 'used'. After compressing the graph, I’ll enumerate all simple paths using DFS, while being mindful of the branching complexity and pruning. Tools like bitmasking will help manage path counting in a graph where cycles and edges may cause significant branching.

Optimizing graph pruning and path counting

To improve path counting, I'll first compress degree-2 vertices and then prune leaves that aren't s or t. By removing unnecessary branches, I'll ensure only relevant paths connect s and t. After processing degree-2 vertices and pruning extraneous subtrees, I'll track distances to key vertices and simplify the graph for easier path enumeration.

Optimizing path pruning and graph traversal

I’ll first compute the 2-core and prune non-essential leaves. Then, using the compressed graph, I’ll remove unnecessary branches while preserving key edges leading to s or t. This will ensure a compact core structure with simple paths. Path enumeration will become straightforward once excess nodes are cleared. Good.

Refining path pruning and enumeration

I’ll compress the graph first, handling cycles and degree-2 nodes. Next, I’ll prune leaves, adjusting degrees without affecting edge weights. Once compression and pruning are complete, I’ll enumerate paths from s to t, using iterative DFS and handling edge cases like self-loops and multi-edges, which simple paths can’t use. Perfect.

Final path pruning and enumeration

I’m implementing the plan. I’ll prune leaves from the compressed graph while keeping track of visited edges. After pruning, I'll enumerate paths using DFS from s to t, taking care not to exceed the path limits. With g ≤ 21, the number of paths will stay manageable, and memory should be sufficient for this operation.

Handling edge cases in path enumeration

I'll handle multi-edges by treating each uniquely in DFS. Self-loops will be skipped, and leaf pruning will reduce unnecessary branches. For the execution, the complexity is manageable with a tree-like structure, and we'll count simple paths using DFS while marking removed nodes. Memory is efficient, and we'll ensure fast output even with large data.

Thought for 3m 44s

回答

下面给出一种适合本题约束（M≤N+20M\le N+20，环路数很少）的做法与 C++ 实现。核心思路是把“绝大多数是链”的图压缩到很小，再在小图上用 DFS 列举所有从 1 到 NN 的简单路径并按长度计数。

思路（中文）

  • 设图的回路数（cyclomatic number）g=M−(N−1)≤21g=M-(N-1)\le 21。
    绝大多数点度数为 2，图“几乎是若干链 + 少量分叉/回边”。

  • 第一步：链压缩（contract degree-2）
    把度数为 2 的非端点（既不是 1 也不是 NN）全部“串联收缩”。
    具体做法：把“关键点”定义为：度≠2\ne 2 或者是 1、NN。从每个关键点沿边走，直到遇到下一个关键点，整段路径合成一条带权边，权重=经过的边数。这样得到一个多重无向图（允许重边），顶点数 KK 约为 O(g)O(g)（严格地说，度≠2\ne 2 的点个数 ≤2g+2\le 2g+2），远小于 NN。

  • 第二步：去叶化（保留 s/t 的 2-core）
    在压缩后的图上，把所有度为 1 且不是 s=1s=1 / t=Nt=N 的点迭代删除（及其相连边）。
    这些“挂在旁边的树枝”在简单 s-ts\text{-}t 路径里不可能出现（因为走进去就出不来而不能重复点），删掉能极大减少分支。
    注意：若原图是一条单纯的路径，此时小图只剩一个 s-ts\text{-}t 的单边（权重为距离），完全没问题。

  • 第三步：在小图上 DFS 枚举简单路径
    小图结点 ≈O(g)\approx O(g)（≤50\le 50 量级），用“按点访问”的 DFS（不允许重复顶点），沿每条边把权重累加成原图中的“边数长度”。到达 tt 时对答案数组 ans[总长度]++。
    由于 g≤21g\le 21，s-ts\text{-}t 简单路径总数至多指数于 gg（大约几百万以内），在 C++ 中完全可承受。
    自环（压缩后可能出现）等价于在同一点绕一圈，会重复顶点，不可能出现在简单路径里，直接忽略。

  • 正确性要点

    • 链压缩保持了任意两关键点之间的所有简单路径长度之和（每条路径长度=若干段链长度之和）。
    • 去叶化不会删除任何可能出现在简单 s-ts\text{-}t 路径上的部分（除了 ss/tt 所在的必要连边）。
    • 在最终小图中，任何简单路径与原图的一条简单路径一一对应，长度为各段权重之和。

  • 复杂度

    • 构建与压缩：O(N+M)O(N+M)。
    • 去叶化：O(K+E′)O(K+E')，K,E′K,E' 为小图的点边数（均为 O(g)O(g) 级别）。
    • DFS：与简单路径条数成正比；在 g≤21g\le 21 下可行。
    • 内存：答案数组大小 NN（最多 2×1052\times 10^5），用 long long 储存计数。

代码（C++，GNU++17）

cpp
#include <bits/stdc++.h>
using namespace std;

struct Edge {
    int u, v;
};
struct CEdge {
    int a, b;
    int w;
    bool alive;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N, M;
    if (!(cin >> N >> M)) return 0;
    vector<Edge> E(M);
    for (int i = 0; i < M; ++i) {
        int u, v; cin >> u >> v;
        --u; --v;
        E[i] = {u, v};
    }

    // 原图：按边索引存邻接
    vector<vector<int>> inc(N);
    vector<int> deg(N, 0);
    for (int i = 0; i < M; ++i) {
        inc[E[i].u].push_back(i);
        inc[E[i].v].push_back(i);
        deg[E[i].u]++; deg[E[i].v]++;
    }
    auto other = [&](int eid, int x) {
        return E[eid].u ^ E[eid].v ^ x;
    };

    // 关键点：度!=2 或者 是 s(0)/t(N-1)
    int s0 = 0, t0 = N - 1;
    vector<char> is_key(N, false);
    for (int v = 0; v < N; ++v) {
        if (deg[v] != 2 || v == s0 || v == t0) is_key[v] = true;
    }

    // 给关键点编号
    vector<int> cid(N, -1);
    int K = 0;
    for (int v = 0; v < N; ++v) if (is_key[v]) cid[v] = K++;

    // 从每个关键点出发，把链收缩为带权边
    vector<char> usedE(M, false);
    vector<CEdge> cedges; cedges.reserve(M);
    for (int u = 0; u < N; ++u) if (is_key[u]) {
        for (int eid : inc[u]) if (!usedE[eid]) {
            usedE[eid] = true;
            int prev = u;
            int cur = other(eid, u);
            int len = 1;
            int last_eid = eid;

            while (!is_key[cur]) {
                // cur 度必为 2，沿另一条边走
                int e0 = inc[cur][0], e1 = inc[cur][1];
                int ne = (e0 == last_eid ? e1 : e0);
                usedE[ne] = true;
                last_eid = ne;
                int nxt = other(ne, cur);
                prev = cur; cur = nxt; len++;
            }
            int a = cid[u], b = cid[cur];
            if (a != b) { // 自环无效，直接丢弃
                cedges.push_back({a, b, len, true});
            }
        }
    }

    if (K == 0) {
        // 所有点都不是关键点？只有可能 N=2 且一条边被自环丢弃的极端，按题意连通不会发生
        // 兜底：输出全 0
        for (int k = 1; k <= N - 1; ++k) cout << 0 << (k==N-1?'\n':' ');
        return 0;
    }

    int s = cid[s0], t = cid[t0];

    // 小图邻接（带边编号）
    int CE = (int)cedges.size();
    vector<vector<pair<int,int>>> adj(K); // (eid, to)
    vector<int> degC(K, 0);
    for (int i = 0; i < CE; ++i) {
        auto [a,b,w,alive] = cedges[i];
        adj[a].push_back({i, b});
        adj[b].push_back({i, a});
        degC[a]++; degC[b]++;
    }

    // 去叶化：删除所有度=1且不是 s/t 的点（迭代）
    vector<char> removed(K, false);
    queue<int> q;
    for (int v = 0; v < K; ++v) {
        if (v == s || v == t) continue;
        if (degC[v] == 1) q.push(v);
    }
    while (!q.empty()) {
        int v = q.front(); q.pop();
        if (removed[v] || v == s || v == t) continue;
        if (degC[v] != 1) continue;
        removed[v] = true;
        for (auto [eid, to] : adj[v]) if (cedges[eid].alive) {
            cedges[eid].alive = false;
            degC[v]--;
            degC[to]--;
            if (to != s && to != t && !removed[to] && degC[to] == 1) q.push(to);
        }
    }

    // DFS 枚举简单路径（按点去重）
    vector<long long> ans(max(1, N), 0LL);
    vector<char> vis(K, false);
    vis[s] = true;

    function<void(int,int)> dfs = [&](int u, int dist) {
        if (u == t) {
            if (dist >= 0 && dist <= N-1) ans[dist]++;
            return;
        }
        for (auto [eid, v] : adj[u]) {
            if (!cedges[eid].alive) continue;
            if (removed[v]) continue;
            if (vis[v]) continue;
            int w = cedges[eid].w;
            if (dist + w > N-1) continue; // 理论上不会超，但保险
            vis[v] = true;
            dfs(v, dist + w);
            vis[v] = false;
        }
    };
    // 若 s 或 t 被完全孤立（无路），DFS 也会给出全 0
    dfs(s, 0);

    // 输出
    for (int k = 1; k <= N - 1; ++k) {
        if (k < (int)ans.size()) cout << ans[k];
        else cout << 0;
        cout << (k == N-1 ? '\n' : ' ');
    }
    return 0;
}

备注

  • 计数使用 long long 足够（在本题 g≤21g\le 21 的范围内，路径总数远小于 2632^{63}）。
  • 若你担心递归，可把 DFS 改成手写栈的迭代版；但小图深度 ≤O(g)\le O(g)（几十层），递归安全。
  • 自环在“简单路径”中永远无效，已直接忽略。
  • 去叶化一定要在链压缩之后做，这样不会误删 s-ts\text{-}t 唯一路径上的中间段。

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