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 + ], res[MAXN + ], path[MAXN + ];
 ll sum[MAXN + ], ans[MAXN + ];
 struct node
 {
     int to, cost;
 };
 vector<node> G[MAXN + ];
 void solve(int x, int d, ll s)
 {
     sum[d] = s;
     path[d] = x;
     int l = lower_bound(sum + , sum + d + , s - a[x]) - sum - ;
     res[path[d]]++;
     res[path[l]]--;
     for (int i = ; i < G[x].size(); i++)
     {
         solve(G[x][i].to, d + , s + G[x][i].cost);
     }
 }
 void dfs(int x)
 {
     ans[x] = res[x];
     for (int i = ; i < G[x].size(); i++)
     {
         dfs(G[x][i].to);
         ans[x] += ans[G[x][i].to];
     }
 }
 int main()
 {
     cin >> n;
     for (int i = ; i <= n; i++)
     {
         scanf("%d", &a[i]);
     }
     for (int i = ; i < n - ; i++)
     {
         int x, y;
         scanf("%d %d", &x, &y);
         node tmp;
         tmp.to = i+;
         tmp.cost = y;
         G[x].push_back(tmp);
     }
     solve(, , );
     dfs();
     for (int i = ; i <= n; i++)
         printf("%I64d ", ans[i]-);
     puts("");
     //("pause");
     return ;
 }