牛客-牛牛染颜色

题目传送门

sol:树形dp，用$dp[u]$表示节点$u$代表的子树合法染色方案的数量，若$u$节点是黑色，则所有儿子随意，产生的方案数为$prod dp[v], v in son[u]$；若$u$节点是白色，则含有黑节点的儿子最多只能有一个，也可以没有。因为$dp[v]$里必有一种方案是整颗$v$子树均为白色，所有产生的方案数为$1 + sum {dp[v] - 1}, v in son[u]$。两者相加即为$dp[u]$。

• 树形dp

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 1000010;
  const int MOD = 1000000007;
  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;
  }
  struct {
      int v, to;
  } edge[2 * MAXN];
  int head[MAXN], total;
  int dp[MAXN], que[MAXN];
  void add_edge(int u, int v) {
      edge[total].v = v;
      edge[total].to = head[u];
      head[u] = total ++;
  }
  void slove(int u) {
      int l = 0, r = -1; 
      que[++r] = u;
      while (l <= r) {
          u = que[l++]; dp[u] = -1;
          for (int i = head[u]; i != -1; i = edge[i].to) {
              int v = edge[i].v;
              if (dp[v] != -1) que[++r] = v;
          }
      }
      while (r >= 0) {
          u = que[r--];
          int tmp1 = 1, tmp2 = 1;
          for (int i = head[u]; i != -1; i = edge[i].to) {
              int v = edge[i].v;
              if (dp[v] == -1) continue;
              tmp1 = 1LL * tmp1 * dp[v] % MOD;
              tmp2 = (tmp2 + dp[v] - 1) % MOD;
          }
          dp[u] = (tmp1 + tmp2) % MOD;
      }
  }
  int main() {
      int n = read();
      memset(head, -1, sizeof(head));
      for (int i = 2; i <= n; i++) {
          int u = read(), v = read();
          add_edge(u, v), add_edge(v, u);
      }
      slove(1);
      printf("%d
  ", dp[1]);
      return 0;
  }

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总结：这道神奇的题目居然卡空间，一开始用vector存图 + dfs递归提交内存超限，改了bfs迭代形式就过了。赛后看了讨论群他们说这是卡vector存图，然鹅我居然在其他地方优化空间。不过这也让我意识到了递归，迭代，前向星和vector在效率上的区别。递归快，迭代省内存，前向星时间空间都优于vector。附一个最开始卡内存的写法。

• 被卡内存的写法

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 1000010;
  const int MOD = 1000000007;
  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 dp[MAXN];
  void dfs(int u, int f) {
      int tmp1 = 1, tmp2 = 1;
      for (int v : edge[u]) {
          if (v == f) continue;
          dfs(v, u);
          tmp1 = 1LL * tmp1 * dp[v] % MOD;
          tmp2 = (tmp2 + dp[v] - 1) % MOD;
      }
      dp[u] = (tmp1 + tmp2) % MOD;
  }
  int main() {
      int n = read();
      for (int i = 2; i <= n; i++) {
          int u = read(), v = read();
          edge[u].push_back(v);
          edge[v].push_back(u);
      }
      dfs(1, -1);
      printf("%d
  ", dp[1]);
      return 0;
  }