UVA 1108-mining Your Own Business

The complete code is given in the Rujia book.

Here's the idea:

There must be one or more double connected components known by test instructions;

Suppose a double-connected component has a cut-top. The Taiping well must not be hit on the top of the cut.

Instead, choose a random point beyond the top of the cut;

Assuming that there is no cut-off, a two well is to be struck on the double connected component

As for the drilling scheme. See Code

#include <cstdio> #include <cstring> #include <vector> #include <stack> #include <map>    using namespace Std;    const int N = 50005;      struct Edge {int u, v;          Edge () {} edge (int u, int v) {this->u = u;      This->v = v;    }  };  int pre[n], bccno[n], Dfs_clock, bcc_cnt;    BOOL Iscut[n];  Vector<int> G[n], bcc[n];    Stack<edge> S;      int dfs_bcc (int u, int fa) {int Lowu = pre[u] = ++dfs_clock;      int child = 0;          for (int i = 0; i < g[u].size (); i++) {int v = g[u][i];          Edge e = Edge (U, v);              if (!pre[v]) {S.push (e);              child++;              int LOWV = DFS_BCC (V, u);              Lowu = min (Lowu, LOWV);                  if (LOWV >= pre[u]) {Iscut[u] = true; bcc_cnt++; Bcc[bcc_cnt].clear (); Start from 1 while (1) {Edge x = S.top ();                      S.pop (); if (BCCNO[X.U]! = bcc_cnt) {bcc[bcc_cnt].push_back (X.U); bccno[x.u] = bcc_cnt;}                      if (BCCNO[X.V]! = bcc_cnt) {bcc[bcc_cnt].push_back (X.V); bccno[x.v] = bcc_cnt;}                  if (x.u = = U && x.v = = v) break;              }}}} else if (Pre[v] < Pre[u] && v! = FA) {s.push (e);          Lowu = min (Lowu, pre[v]);      }} if (FA < 0 && child = = 1) Iscut[u] = false;  return LOWU;    } int St;      void Find_bcc () {memset (pre, 0, sizeof (pre));      memset (iscut, 0, sizeof (iscut));      memset (bccno, 0, sizeof (BCCNO));      Dfs_clock = bcc_cnt = 0;  DFS_BCC (0,-1);    } int n, m;    typedef long Long LL;      void Solve () {ll ans1 = 0, ans2 = 1;          for (int i = 1; I <= bcc_cnt; i++) {int cut_cnt = 0;          for (int j = 0; J < Bcc[i].size (); j + +) if (Iscut[bcc[i][j]]) cut_cnt++;         if (cut_cnt = = 1) {ans1++;      Ans2 *= (LL) (Bcc[i].size ()-cut_cnt);           }} if (bcc_cnt = = 1) {ans1 = 2;      Ans2 = (ll) bcc[1].size () * (Bcc[1].size ()-1)/2;  } printf ("%lld%lld\n", ans1, ANS2);      } int main () {int cas = 0;          while (~SCANF ("%d", &m) && m) {int u, V, Max = 0;              while (m--) {scanf ("%d%d", &u, &v); u--;              v--;              G[u].push_back (v);              G[v].push_back (U);              max = max (max, u);          max = max (max, v);          } find_bcc ();          printf ("Case%d:", ++cas);          Solve ();      for (int i = 0; I <= Max; i++) g[i].clear ();  } return 0;   }

UVA 1108-mining Your Own Business