题目链接

题目

题目描述

Farmer John is decorating his Spring Equinox Tree (like a Christmas tree but popular about three months later). It can be modeled as a rooted mathematical tree with N (1 <= N <= 100,000) elements, labeled 1...N, with element 1 as the root of the tree. Each tree element e > 1 has a parent, PeP_ePe (1 <= \(P_e\) <= N). Element 1 has no parent (denoted '-1' in the input), of course, because it is the root of the tree.

Each element i has a corresponding subtree (potentially of size 1) rooted there. FJ would like to make sure that the subtree corresponding to element i has a total of at least \(C_i\) (0 <= \(C_i\) <= 10,000,000) ornaments scattered among its members. He would also like to minimize the total amount of time it takes him to place all the ornaments (it takes time K*\(T_i\) to place K ornaments at element i (1 <= \(T_i\)​ <= 100)).

Help FJ determine the minimum amount of time it takes to place ornaments that satisfy the constraints. Note that this answer might not fit into a 32-bit integer, but it will fit into a signed 64-bit integer.

For example, consider the tree below where nodes located higher on
the display are parents of connected lower nodes (1 is the root): 1
|
2
|
5
/ \
4 3 Suppose that FJ has the following subtree constraints: Minimum ornaments the subtree requires
| Time to install an ornament
Subtree | |
root | C_i | T_i
--------+--------+-------
1 | 9 | 3
2 | 2 | 2
3 | 3 | 2
4 | 1 | 4
5 | 3 | 3 Then FJ can place all the ornaments as shown below, for a total
cost of 20: 1 [0/9(0)] legend: element# [ornaments here/ | total ornaments in subtree(node install time)]
2 [3/9(6)]
|
5 [0/6(0)]
/ \
[1/1(4)] 4 3 [5/5(10)]

输入描述

  • Line 1: A single integer: N
  • Lines 2..N+1: Line i+1 contains three space-separated integers: PiP_iPi, CiC_iCi, and TiT_iTi

输出描述

  • Line 1: A single integer: The minimum time to place all the ornaments

示例1

输入

5
-1 9 3
1 2 2
5 3 2
5 1 4
2 3 3

输出

20

题解

知识点:贪心,DFS,树形dp。

每个节点为根的子树都有一个最小要求的装饰数量,显然叶子节点只能全挂上去,随后向上考虑。对于一个子树,肯定把装饰挂在花费最小的节点上,因此可以回溯同时更新子树最小值,同时还需要一个记录已经挂了多少的数组。

注意结果可能超 int

时间复杂度 \(O(n)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>
#define ll long long using namespace std; struct edge {
int to, nxt;
}e[100007];
int h[100007], cnt;
int root, c[100007], t[100007];///某节点的需求装饰;某子树所有节点的最小t;
ll ans, csum[100007]; ///某子树已有装饰 void add(int u, int v) {
e[cnt].to = v;
e[cnt].nxt = h[u];
h[u] = cnt++;
} void dfs(int u) {
if (!~h[u]) {
ans += c[u] * t[u];
csum[u] = c[u];
return;
}
for (int i = h[u];~i;i = e[i].nxt) {
int v = e[i].to;
dfs(v);
csum[u] += csum[v];
t[u] = min(t[u], t[v]);
}
ans += max(c[u] - csum[u], 0LL) * t[u];
csum[u] = max(csum[u], 0LL + c[u]);
} int main() {
std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
memset(h, -1, sizeof(h));
int n;
cin >> n;
for (int i = 1;i <= n;i++) {
int p;
cin >> p >> c[i] >> t[i];
if (p == -1) root = i;
else add(p, i);
}
dfs(root);
cout << ans << '\n';
return 0;
}

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