UVA 1108 - Mining Your Own Business

標籤：

劉汝佳書上都給出了完整的代碼

在這裡理一下思路：

由題意知肯定存在一個或者多個雙連通分量；

假設某一個雙連通分量有割頂。那太平井一定不能打在割頂上。

而是選擇割頂之外的隨意一個點；

假設沒有割頂，則要在該雙連通分量上打兩個井

至於打井方案。見代碼

#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