ICPC Central Europe Regional Contest 2019ISaba1000kg Jathon

题目传送门

sol:因为查询的点一共最多只有$100000$个。若当本次查询的点比较少，小于$\sqrt{100000}$，那我们可以枚举起始点$u$和终点$v$，若$u, v$之间有边，则用并查集将$u, v$节点合并，最后统计有多少个联通块；若当本次查询的点比较多，大于$\sqrt{100000}$，那我们可以枚举所有边，若$u, v$都在本次查询的点集内，则用并查集将$u, v$节点合并，最后统计有多少个联通块。

• 二分+并查集

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 100010;
  inline int read() {
      int n = 0, f = 1; char c = getchar();
      while (c < '0' || c > '9') {
          if (c == '-') f = -f;
          c = getchar();
      }
      while (c >= '0' && c <= '9') {
          n = 10 * n + (c ^ '0');
          c = getchar();
      }
      return f * n;
  }
  vector<int> edge[MAXN];
  int node[MAXN], dsu[MAXN];
  int find(int i) {
      if (dsu[i] == -1) return i;
      return dsu[i] = find(dsu[i]);
  }
  bool search_1(int u, int v) {
      int l = -1, r = edge[u].size();
      while (l < r - 1) {
          int m = l + r >> 1;
          if (edge[u][m] == v) return true;
          if (edge[u][m] < v) l = m;
          if (edge[u][m] > v) r = m;
      }
      return false;
  }
  bool search_2(int l, int r, int k) {
      if (node[l] == k || node[r] == k) return true;
      while (l < r - 1) {
          int m = l + r >> 1;
          if (node[m] == k) return true;
          if (node[m] < k) l = m;
          if (node[m] > k) r = m;
      }
      return false;
  }
  void slove_1(int n, int k) {
      for (int i = 1; i <= k; i++) {
          for (int j = 1; j < i; j++) {
              if (search_1(node[i], node[j])) {
                  int fu = find(node[i]);
                  int fv = find(node[j]);
                  if (fu != fv) dsu[fu] = fv;
              }
          }
      }
      int ans = 0;
      for (int i = 1; i <= k; i++) {
          if (dsu[node[i]] == -1) ans ++;
          else dsu[node[i]] = -1;
      }
      printf("%d\n", ans);
  }
  int slove_2(int n, int k) {
      for (int u = 1; u <= n; u++) {
          for (int v : edge[u]) {
              if (search_2(1, k, u) && search_2(1, k, v)) {
                  int fu = find(u);
                  int fv = find(v);
                  if (fu != fv) dsu[fu] = fv;
              }
          }
      }
      int ans = 0;
      for (int i = 1; i <= k; i++) {
          if (dsu[node[i]] == -1) ans ++;
          else dsu[node[i]] = -1;
      }
      printf("%d\n", ans);
  }
  int main() {
      int n = read(), m = read(), q = read();
      for (int i = 1; i <= m; i++) {
          int u = read(), v = read();
          edge[u].push_back(v);
          edge[v].push_back(u);
      }
      for (int i = 1; i <= n; i++) {
          sort(edge[i].begin(), edge[i].end());
      }
      memset(dsu, -1, sizeof(dsu));
      while (q--) {
          int k = read();
          for (int i = 1; i <= k; i++) node[i] = read();
          sort(node + 1, node + 1 + k);
          if (k <= 320) slove_1(n, k);
          else slove_2(n, k);
      }
      return 0;
  }

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  总结：这道题的解法真妙，一道题里用到了两种不同策略。$solve1$的复杂度是$O(k^2\log_2k)$，当k比较小时非常快，$slove2$的复杂度是$O(m\log_2k)$，随着$k$的增加变化不大。所以$k$小用$slove1$，$k$大用$slove2$。