题目
题目描述
Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place.
Each cow lives in one of N (1 <= N <= 100,000) different barns (conveniently numbered 1..N) which are connected by N-1 roads in such a way that it is possible to get from any barn to any other barn via the roads. Road i connects barns \(A_i\) and \(B_i\) (1 <= \(A_i\) <= N; 1 <= \(B_i\) <= N) and has length \(L_i\) (1 <= \(L_i\) <= 1,000). The Great Cow Gathering can be held at any one of these N barns. Moreover, barn i has \(C_i\) (0 <= \(C_i\) <= 1,000) cows living in it.
When choosing the barn in which to hold the Cow Gathering, Bessie wishes to maximize the convenience (which is to say minimize the inconvenience) of the chosen location. The inconvenience of choosing barn X for the gathering is the sum of the distances all of the cows need to travel to reach barn X (i.e., if the distance from barn i to barn X is 20, then the travel distance is \(C_i*20\)). Help Bessie choose the most convenient location for the Great Cow Gathering.
Consider a country with five barns with [various capacities] connected by various roads of varying lengths. In this set of barns, neither barn 3 nor barn 4 houses any cows.
1 3 4 5
@--1--@--3--@--3--@[2]
[1] |
2
|
@[1]
2
Bessie can hold the Gathering in any of five barns; here is the table of inconveniences calculated for each possible location: Gather ----- Inconvenience ------
Location B1 B2 B3 B4 B5 Total
1 0 3 0 0 14 17
2 3 0 0 0 16 19
3 1 2 0 0 12 15
4 4 5 0 0 6 15
5 7 8 0 0 0 15
If Bessie holds the gathering in barn 1, then the inconveniences from each barn are:
Barn 1 0 -- no travel time there!
Barn 2 3 -- total travel distance is 2+1=3 x 1 cow = 3
Barn 3 0 -- no cows there!
Barn 4 0 -- no cows there!
Barn 5 14 -- total travel distance is 3+3+1=7 x 2 cows = 14
So the total inconvenience is 17.
The best possible convenience is 15, achievable at by holding the
Gathering at barns 3, 4, or 5.
输入描述
- Line 1: A single integer: N
- Lines 2..N+1: Line i+1 contains a single integer: \(C_i\)
- Lines N+2..2*N: Line i+N+1 contains three integers: \(A_i\), \(B_i\), and \(L_i\)
输出描述
- Line 1: The minimum inconvenience possible
示例1
输入
5
1
1
0
0
2
1 3 1
2 3 2
3 4 3
4 5 3
输出
15
题解
知识点：树形dp。
题目给了一个树，树边有边权表示距离，节点有点权表示牛的数量，现在要选一个树上点，使得所有牛到这个点的距离最小。因此，这是一个二次扫描+换根dp的问题，因为是解决每个点对于整棵树的问题。
第一遍dp，先以 \(1\) 为根，得出节点 \(1\) 的答案（当然可选其他点作为根）。设 \(dp[u]\) 为从 \(u\) 开始到以 \(u\) 为根的子树中各个节点牛的总距离。转移方程为：
\[dp[u] = \sum (dp[v] + sz[v] \cdot w) \]第二遍dp，从 \(1\) 出发，处理所有节点的答案。设 \(dp'[u]\) 为从 \(u\) 到牛的总距离，则有转移方程：
\[dp'[v] = dp'[u] + (sz[1] - 2 \cdot sz[v]) \]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;
int a[100007];
vector<pair<int, int>> g[100007];
ll sz[100007], dp[100007];
void dfs1(int u, int fa) {
sz[u] = a[u];
for (auto [v, w] : g[u]) {
if (v == fa) continue;
dfs1(v, u);
sz[u] += sz[v];
dp[u] += dp[v] + sz[v] * w;
}
}
void dfs2(int u, int fa) {
for (auto [v, w] : g[u]) {
if (v == fa) continue;
dp[v] = dp[u] + (sz[1] - 2 * sz[v]) * w;
dfs2(v, u);
}
}
int main() {
std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
cin >> n;
for (int i = 1;i <= n;i++) cin >> a[i];
for (int i = 1;i < n;i++) {
int u, v, w;
cin >> u >> v >> w;
g[u].push_back({ v,w });
g[v].push_back({ u,w });
}
dfs1(1, 0);
dfs2(1, 0);
//for (int i = 1;i <= n;i++) cout << dp[i] << ' ';
ll ans = ~(1LL << 63);
for (int i = 1;i <= n;i++) ans = min(ans, dp[i]);
cout << ans << '\n';
return 0;
}