Luogu5058 [ZJOI2004]嗅探器

\(\verb|Luogu5058 [ZJOI2004]嗅探器|\)

给定一张 \(n\) 个点， \(m\) 条边的无向图，和两点 \(s,\ t\) ，求 \(s\to t\) 编号最小的必经点（排除 \(s,\ t\) ）

\(n\leq100\)

tarjan

显然只需缩点后的树上 \(bl_s\) 到 \(bl_t\) 上找答案，统计割点贡献即可

然而此题有更简单的做法……

从 \(s\) 开始 tarjan，点 \(u\) 对答案有贡献当且仅当满足以下四个条件：

1. \(u\neq s,\ t\)
2. \(cut_u=\operatorname{true}\)
3. \(dfn_v\leq dfn_t\) ，因为终点必须在 \(u\) 之后访问到
4. \(dfn_u\leq low_t\) ，因为路径必须要经过 \(u\) 点

然后上板子

时间复杂度 \(O(n+m)\)

这份代码是缩点后统计链的……

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

const int maxn = 110, inf = INT_MAX;
bool cut[maxn];
int n, m, A, B, h[maxn], q[maxn << 1], pre[maxn << 1];
int top, dcc, tot, st[maxn], bl[maxn], dfn[maxn], low[maxn];

struct edges {
  int nxt, to;
  edges(int x = 0, int y = 0) :
    nxt(x), to(y) {}
} e[maxn * maxn * 2];

vector <int> E[maxn << 1], d[maxn];

void addline(int u, int v) {
  static int cnt;
  e[++cnt] = edges(h[u], v), h[u] = cnt;
}

void tarjan(int u, int f) {
  static int now;
  st[++top] = u;
  dfn[u] = low[u] = ++now;
  if (!f && !h[u]) {
    d[++dcc].push_back(u);
    return;
  }
  for (int i = h[u], chd = 0; i; i = e[i].nxt) {
    int v = e[i].to;
    if (!dfn[v]) {
      tarjan(v, 1);
      low[u] = min(low[u], low[v]);
      if (dfn[u] <= low[v]) {
        cut[u] |= f || chd++;
        for (dcc++; st[top + 1] != v; top--) {
          d[dcc].push_back(st[top]);
        }
        d[dcc].push_back(u);
      }
    } else {
      low[u] = min(low[u], dfn[v]);
    }
  }
}

int bfs(int S, int T) {
  if (S == T) return inf;
  int l = 1, r = 1;
  q[1] = S, pre[S] = -1;
  while (l <= r) {
    int u = q[l++];
    for (int v : E[u]) {
      if (!pre[v]) {
        q[++r] = v, pre[v] = u;
      }
    }
  }
  int res = inf;
  for (int u = pre[T]; u != S; u = pre[u]) {
    if (u > dcc) res = min(res, u);
  }
  return res;
}

int main() {
  scanf("%d", &n);
  int u, v;
  while (scanf("%d %d", &u, &v) && u && v) {
    addline(u, v), addline(v, u);
  }
  scanf("%d %d", &A, &B);
  for (int i = 1; i <= n; i++) {
    if (!dfn[i]) tarjan(i, 0);
  }
  tot = dcc;
  for (int i = 1; i <= n; i++) {
    if (cut[i]) bl[i] = ++tot;
  }
  for (int i = 1; i <= dcc; i++) {
    for (int j = 0, _sz = int(d[i].size()); j < _sz; j++) {
      int x = d[i][j];
      if (cut[x]) {
        E[i].push_back(bl[x]);
        E[bl[x]].push_back(i);
      } else {
        bl[x] = i;
      }
    }
  }
  int res = bfs(bl[A], bl[B]);
  if (res > 1e9) {
    puts("No solution");
  } else {
    for (int i = 1; i <= n; i++) {
      if (cut[i] && bl[i] == res) {
        printf("%d", i); break;
      }
    }
  }
  return 0;
}

原文地址：https://www.cnblogs.com/Juanzhang/p/10660452.html