「考前日志」11.16

总结

最近状态不对劲

今日已完成

• 一节文化课：英语完形填空小技巧

  =_=这节课讲的东西还不如我自己的方法

• AcWing287 积蓄程度
  换根DP，树形DP

  挺基础的换根DP了。随便找一个点作为根，记录每个点的度数，设 \(f_x\) 表示以 \(x\) 为根的子树，把 \(x\) 作为源点，从 \(x\) 出发流向子树的最大流量是多少，用 \(du_x\) 表示 \(x\) 的度数，有转移方程：

  \[f_{x}=\sum\limits_{to\in{Son_{x}}}\begin{cases}\min(f_{to},val(x,to))&\text{if }du_{to}>1\\val(x,to)&\text{if }du_{to}=1\end{cases} \]

  之后考虑换根求出以每个点为根的答案，设 \(g_x\) 表示以 \(x\) 为整棵树的根所能流过的最大流量，因为每个 \(x\) 的儿子 \(to\) 都已经处理完了自身子树内的最大流量 \(f_{to}\)，所以只需要处理 \(x\) 除了以 \(to\) 为根的子树之外的部分对 \(g_{to}\) 的贡献，那么有：

  \[g_{to}=f_{to}+\begin{cases}\min(g_{x}-\min({f_{to},val(x,to)}),val(x,to))&\text{if }du_{x}>1\\val(x,to)&\text{if } du_{x}=1\end{cases} \]

  发现是要自顶向下更新的，所以先更新，再进一步遍历。

  调了挺久发现是数组开小了= =

  

  #include <cmath>
  #include <queue>
  #include <cstdio>
  #include <vector>
  #include <cstring>
  #include <iostream>
  #include <algorithm>
  #define int long long
  using namespace std;
  
  const int A = 2e5 + 11;
  const int B = 1e6 + 11;
  const int mod = 1e9 + 7;
  const int inf = 0x3f3f3f3f;
  
  inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return x * f;
  }
  
  struct node { int to, nxt, val; } e[A << 1];
  int n, cnt, head[A], f[A], vis[A], g[A], du[A];
  
  inline void add(int from, int to, int val) {
    e[++cnt].to = to;
    e[cnt].val = val;
    e[cnt].nxt = head[from];
    head[from] = cnt;
  }
  
  inline void init() {
    cnt = 0;
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    memset(du, 0, sizeof(du));
    memset(head, 0, sizeof(head));
  }
  
  void dfs(int x, int fa) {
    for (int i = head[x]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (to == fa) continue;
      dfs(to, x);
      if (du[to] == 1) f[x] = f[x] + e[i].val;
      else f[x] = f[x] + min(e[i].val, f[to]);
    }
  }
  
  void change(int x, int fa) {
    for (int i = head[x]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (to == fa) continue;
      if (du[x] == 1) g[to] = f[to] + e[i].val;
      else g[to] = f[to] + min(e[i].val, g[x] - min(f[to], e[i].val));
      change(to, x);
    }
  }
  
  inline void solve() {
    init();
    n = read();
    for (int i = 1; i < n; i++) {
      int x = read(), y = read(), val = read();
      du[x]++, du[y]++;
      add(x, y, val), add(y, x, val);
    }
    dfs(1, 0);
    g[1] = f[1];
    change(1, 0);
    int ans = 0;
    for (int i = 1; i <= n; i++) ans = max(ans, g[i]);
    cout << ans << '\n';
    return;
  }
  
  signed main() {
    int T = read();
    while (T--) solve();
  }
  

• 洛谷 P2018 消息传递

  树形DP，见此博客

  #include <cmath>
  #include <queue>
  #include <cstdio>
  #include <vector>
  #include <cstring>
  #include <iostream>
  #include <algorithm>
  using namespace std;
  
  const int A = 1e3 + 11;
  const int B = 1e6 + 11;
  const int mod = 1e9 + 7;
  const int inf = 0x3f3f3f3f;
  
  inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return x * f;
  }
  
  struct node { int to, nxt; } e[A << 1];
  int n, ans[A], f[A], head[A], cnt = 0, res = inf; 
  
  inline void add(int from, int to) {
    e[++cnt].to = to;
    e[cnt].nxt = head[from];
    head[from] = cnt;
  }
  
  bool cmp(int x, int y) {
    return x > y;
  }
  
  inline void dfs(int x, int fa) {
    int tot = 0, b[1000] = {0};
    for (int i = head[x]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (to == fa) continue;
      dfs(to, x);
      b[++tot] = f[to];
    }
    sort(b + 1, b + 1 + tot, cmp);
    for (int i = 1; i <= tot; i++) 
      f[x] = max(f[x], b[i] + i);
  }
  
  int main() {
    n = read();
    for (int i = 2; i <= n; i++) {
      int x = read();
      add(x, i), add(i, x);
    }
    for (int i = 1; i <= n; i++) {
      memset(f, 0, sizeof(f));
      dfs(i, 0);
      res = min(f[i], res);
      ans[i] = f[i];
    }
    cout << res + 1 << '\n';
    for (int i = 1; i <= n; i++) 
      if (ans[i] == res) cout << i << " ";
    puts("");
    return 0;
  }
  

• AcWing288 休息时间

  环形DP问题的转化

  #include <cmath>
  #include <queue>
  #include <cstdio>
  #include <vector>
  #include <cstring>
  #include <iostream>
  #include <algorithm>
  using namespace std;
  
  const int A = 3831;
  const int B = 1e6 + 11;
  const int mod = 1e9 + 7;
  const int inf = 0x3f3f3f3f;
  
  inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return x * f;
  }
  
  int n, m, a[A], f[2][A][2], ans, now = 0, last = 1;
  
  int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    memset(f, 0xcf, sizeof(f));
    f[now][0][0] = f[now][1][1] = 0;
    for (int i = 2; i <= n; i++) {
      swap(now, last);
      //debug：j从1开始枚举的 
      for (int j = 0; j <= m; j++) {
        f[now][j][0] = max(f[last][j][0], f[last][j][1]);
        if (j) f[now][j][1] = max(f[last][j - 1][0], f[last][j - 1][1] + a[i]);
      }
      memset(f[last], 0xcf, sizeof(f[last]));
    }
    ans = max(f[now][m][0], f[now][m][1]);
    memset(f, 0xcf, sizeof(f));
    f[now][1][1] = a[1], f[now][0][0] = 0;
    //忘记初始化f[now][0][0] 
    for (int i = 2; i <= n; i++) {
      swap(now, last);
      //debug：j从1开始枚举的
      for (int j = 0; j <= m; j++) {
        f[now][j][0] = max(f[last][j][0], f[last][j][1]);
        if (j) f[now][j][1] = max(f[last][j - 1][0], f[last][j - 1][1] + a[i]);
      }
      memset(f[last], 0xcf, sizeof(f[last]));
    }
    ans = max(ans, f[now][m][1]);
    cout << ans << '\n';
    return 0;
  }