CF739B Alyona and a tree

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前缀和+倍增+树上差分

假设\(v\)是\(u\)子树中的一个点，那么\(u\)能控制\(v\)的条件是受\(v\)的权值的限制，而并非\(u\)。因此我们就能想到计算每一个点的贡献，即\(v\)有多少个祖先能控制它。这样就能想到暴力的做法：枚举每一个点\(i\)，向上爬直到两点间距离大于\(a_i\)为止。然后树上差分（准确说是链上差分）即可。至于两点间距离，采用前缀和相减。

但这样的复杂度能达到\(O(n^2)\)，因此我们可以用倍增优化一步步向上跳，达到\(O(nlogn)\)。

总结一下，先\(dfs\)一遍求出每一个点到根节点的距离和差分数组，复杂度\(O(nlogn)\)；然后对于每一个点倍增向上跳，并修改差分数组，复杂度也是\(O(nlogn)\)；最后\(O(n)\) \(dfs\)一遍求查差分组的树上前缀和。

#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 rg register
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;
ll a[maxn];
struct Edge
{
  int nxt, to, w;
}e[maxn];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w};
  head[x] = ecnt;
}

int fa[21][maxn];
ll dis[maxn];
void dfs(int now)
{
  for(int i = 1; i <= 20; ++i)
    fa[i][now] = fa[i - 1][fa[i - 1][now]];
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      fa[0][v] = now;
      dis[v] = dis[now] + e[i].w;
      dfs(v);
    }
}

int dif[maxn];
void solve(int now)
{
  int x = now;
  for(int i = 20; i >= 0; --i)
      if(fa[i][x] && dis[now] - dis[fa[i][x]] <= a[now]) x = fa[i][x];
  if(x != 1) dif[fa[0][x]]--;
  if(now != 1) dif[fa[0][now]]++;
}

void dfs2(int now)
{
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      dfs2(v);
      dif[now] += dif[v];
    }
}

int main()
{
  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(), w = read();
      addEdge(x, i, w);
    }
  dfs(1);
  for(int i = 1; i <= n; ++i) solve(i);
  dfs2(1);
  for(int i = 1; i <= n; ++i) write(dif[i]), space; enter;
  return 0;
}