CF739B

深搜的过程中保存路径，二分路径中满足要求的区段。不必将每个节点的ans加1，只需将合法区段末尾加1同时将开头减1来表示并保存在一个“前缀”数组中即可。最后再dfs一次累加得到答案。

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#define MAXN 200000
using namespace std;
typedef long long ll;
int n, a[MAXN + 5], res[MAXN + 5], path[MAXN + 5];
ll sum[MAXN + 5], ans[MAXN + 5];
struct node
{
    int to, cost;
};
vector<node> G[MAXN + 5];
void solve(int x, int d, ll s)
{
    sum[d] = s;
    path[d] = x;
    int l = lower_bound(sum + 1, sum + d + 1, s - a[x]) - sum - 1;
    res[path[d]]++;
    res[path[l]]--;
    for (int i = 0; i < G[x].size(); i++)
    {
        solve(G[x][i].to, d + 1, s + G[x][i].cost);
    }
}
void dfs(int x)
{
    ans[x] = res[x];
    for (int i = 0; i < G[x].size(); i++)
    {
        dfs(G[x][i].to);
        ans[x] += ans[G[x][i].to];
    }
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < n - 1; i++)
    {
        int x, y;
        scanf("%d %d", &x, &y);
        node tmp;
        tmp.to = i+2;
        tmp.cost = y;
        G[x].push_back(tmp);
    }
    solve(1, 1, 0);
    dfs(1);
    for (int i = 1; i <= n; i++)
        printf("%I64d ", ans[i]-1);
    puts("");
    //("pause");
    return 0;
}