嘟嘟嘟


這題真沒想到這麽簡單……


首先有60分大禮：\(O(n ^ 2logn)\)貪心。（我也不知道為啥就是對的）
然後又送15分鏈：維護兩個堆，每次去堆頂的最大值。


這時候得到75分已經很開心了，但其實離AC也就差一點點。
鏈的做法已經給了我們提示：合並兩個堆。其實這就相當於二叉樹。那多叉樹呢？就合並多個堆唄！從子樹向上遞歸的時候不斷將子樹啟發式合並，復雜度就保證了。


代碼真的很短

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, a[maxn];
struct Edge
{
  int nxt, to;
}e[maxn];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
  e[++ecnt] = (Edge){head[x], y};
  head[x] = ecnt;
}

priority_queue<int> q[maxn];
int siz[maxn], tp[maxn];
In int dfs(int now)
{
  int id1 = now;
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      v = e[i].to;
      int id2 = dfs(v);
      if(q[id1].size() < q[id2].size()) swap(id1, id2);
      int Siz = q[id2].size();
      for(int j = 1; j <= Siz; ++j) tp[j] = q[id1].top(), q[id1].pop();
      for(int j = 1; j <= Siz; ++j)
    {
      q[id1].push(max(tp[j], q[id2].top()));
      q[id2].pop();
    }
    }
  q[id1].push(a[now]);
  return id1;
}

int main()
{
  Mem(head, -1);
  n = read();
  for(int i = 1; i <= n; ++i) a[i] = read();
  for(int i = 2, x; i <= n; ++i) x = read(), addEdge(x, i);
  int id = dfs(1);
  ll ans = 0;
  while(!q[id].empty()) ans += q[id].top(), q[id].pop();
  write(ans), enter;
  return 0;
}
 

這題我考場上寫了75分，結果成績一出只剩20分了，當時真的懷疑人生，因為我對自己的暴力有十足的信心。然後查代碼的時候看到了觸目驚心的一幕：

if(judge_line) {work1(); return 0;}

天知道我judge_line()後面的括號去哪了。
-Wall沒給我報錯，自己復查的時候也沒發現，編譯也能過，就這樣白白沒了55分。
考場代碼全放上來吧

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
const int N = 18;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), las = ' ';
  while(!isdigit(ch)) las = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(las == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) putchar('-'), x = -x;
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen("spring.in", "r", stdin);
  freopen("spring.out", "w", stdout);
#endif
}

int n, a[maxn];
struct Edge
{
  int nxt, to;
}e[maxn];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
  e[++ecnt] = (Edge){head[x], y};
  head[x] = ecnt;
}

int fa[N + 2][maxn], dep[maxn], siz[maxn];
In void dfs(int now, int _f)
{
  siz[now] = 1;
  for(int i = 1; (1 << i) <= dep[now]; ++i)
    fa[i][now] = fa[i - 1][fa[i - 1][now]];
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      dep[v] = dep[now] + 1, fa[0][v] = now;
      dfs(v, now);
      siz[now] += siz[v];
    }
}
In int lca(int x, int y)
{
  if(dep[x] < dep[y]) swap(x, y);
  for(int i = N; i >= 0; --i)
    if(dep[x] - (1 << i) >= dep[y]) x = fa[i][x];
  if(x == y) return x;
  for(int i = N; i >= 0; --i)
    if(fa[i][x] ^ fa[i][y]) x = fa[i][x], y = fa[i][y];
  return fa[0][x];
}

struct Node
{
  int val, siz, id;
  In bool operator < (const Node& oth)const
  {
    return val > oth.val || (val == oth.val && siz < oth.siz);
  }
}t[maxn];
vector<int> vec[maxn];
int cnt = 0;
ll ans = 0;
In void work0()
{
  for(int i = 1; i <= n; ++i) t[i] = (Node){a[i], siz[i], i};
  sort(t + 1, t + n + 1);
  for(int i = 1; i <= n; ++i)
    {
      bool flg1 = 0;
      for(int j = 1; j <= cnt && !flg1; ++j)
    {
      bool flg2 = 1;
      for(int k = 0; k < (int)vec[j].size() && flg2; ++k)
        {
          int v = vec[j][k], z = lca(t[i].id, v);
          if(z == v || z == t[i].id) flg2 = 0;
        }
      if(flg2) vec[j].push_back(t[i].id), flg1 = 1;
    }
      if(!flg1) vec[++cnt].push_back(t[i].id), ans += t[i].val;
    }
  write(ans), enter;
}

int du[maxn];
In bool judge_line()
{
  for(int i = 1; i <= n; ++i) if(du[i] > 2) return 0;
  return 1;
}
priority_queue<int> q[3];
In void dfs1(int now, int _f, int pos)
{
  q[pos].push(a[now]);
  for(int i = head[now]; ~i; i = e[i].nxt) 
    dfs1(e[i].to, now, pos);
}
In void work1()
{
  for(int i = head[1], j = 1; ~i; i = e[i].nxt, ++j)
    dfs1(e[i].to, 1, j);
  while(!q[1].empty() || !q[2].empty())
    {
      if(q[2].empty()) {ans += q[1].top(), q[1].pop(); continue;}
      if(q[1].empty()) {ans += q[2].top(), q[2].pop(); continue;}
      if(q[1].top() >= q[2].top()) ans += q[1].top(); 
      else ans += q[2].top();
      q[1].pop(), q[2].pop();
    }
  write(ans + a[1]), enter;
}

int main()
{
  MYFILE();
  Mem(head, -1);
  n = read();
  for(int i = 1; i <= n; ++i) a[i] = read();
  for(int i = 2; i <= n; ++i)
    {
      int x = read();
      addEdge(x, i); 
      ++du[x], ++du[i];
    }
  if(judge_line) {work1(); return 0;}
  dfs(1, 0);
  if(n <= 2000) {work0(); return 0;}
  work0();
  return 0;
}
/*
8 
1 5 2 3 4 1 4 2
1 2 1 3 4 5 7
 */
 

[十二省聯考2019]D2T2春節十二響