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。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <algorithm>
 5 #define MAXN 200000
 6 using namespace std;
 7 typedef long long ll;
 8 int n, a[MAXN + 5], res[MAXN + 5], path[MAXN + 5];
 9 ll sum[MAXN + 5], ans[MAXN + 5];
10 struct node
11 {
12     int to, cost;
13 };
14 vector<node> G[MAXN + 5];
15 void solve(int x, int d, ll s)
16 {
17     sum[d] = s;
18     path[d] = x;
19     int l = lower_bound(sum + 1, sum + d + 1, s - a[x]) - sum - 1;
20     res[path[d]]++;
21     res[path[l]]--;
22     for (int i = 0; i < G[x].size(); i++)
23     {
24         solve(G[x][i].to, d + 1, s + G[x][i].cost);
25     }
26 }
27 void dfs(int x)
28 {
29     ans[x] = res[x];
30     for (int i = 0; i < G[x].size(); i++)
31     {
32         dfs(G[x][i].to);
33         ans[x] += ans[G[x][i].to];
34     }
35 }
36 int main()
37 {
38     cin >> n;
39     for (int i = 1; i <= n; i++)
40     {
41         scanf("%d", &a[i]);
42     }
43     for (int i = 0; i < n - 1; i++)
44     {
45         int x, y;
46         scanf("%d %d", &x, &y);
47         node tmp;
48         tmp.to = i+2;
49         tmp.cost = y;
50         G[x].push_back(tmp);
51     }
52     solve(1, 1, 0);
53     dfs(1);
54     for (int i = 1; i <= n; i++)
55         printf("%I64d ", ans[i]-1);
56     puts("");
57     //("pause");
58     return 0;
59 }