F. Tree with Maximum Cost time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output
You are given a tree consisting exactly of n vertices. Tree is a connected undirected graph with n−1 edges. Each vertex v of this tree has a value av assigned to it.
Let dist(x,y) be the distance between the vertices x and y. The distance between the vertices is the number of edges on the simple path between them.
Let’s define the cost of the tree as the following value: firstly, let’s fix some vertex of the tree. Let it be v. Then the cost of the tree is ∑ i = 1 n \sum_{i=1}^n∑
i=1
n
dist(i,v)⋅ai
Your task is to calculate the maximum possible cost of the tree if you can choose v arbitrarily.
Input
The first line contains one integer n, the number of vertices in the tree (1≤n≤2⋅105).
The second line of the input contains n integers a1,a2,…,an (1≤ai≤2⋅105), where ai is the value of the vertex i.
Each of the next n−1 lines describes an edge of the tree. Edge i is denoted by two integers ui and vi, the labels of vertices it connects (1≤ui,vi≤n, ui≠vi).
It is guaranteed that the given edges form a tree.
Output
Print one integer — the maximum possible cost of the tree if you can choose any vertex as v.
Examples
input 1
8 9 4 1 7 10 1 6 5 1 2 2 3 1 4 1 5 5 6 5 7 5 8
output 1
121
input 2
1 1337
output 2
0
Note
Picture corresponding to the first example:
You can choose the vertex 3 as a root, then the answer will be 2⋅9+1⋅4+0⋅1+3⋅7+3⋅10+4⋅1+4⋅6+4⋅5=18+4+0+21+30+4+24+20=121.
In the second example tree consists only of one vertex so the answer is always 0.
题目大意：
给出一个具有n个点的无向图，给出 n-1条边，然后求出如果以一个点为树的根，那么这棵树的Cost是多少，Cost定义为 ∑ i = 1 n \sum_{i=1}^n∑
i=1
n
dist(i,v)⋅ai，dist(a,b)的含义就是两个点之间的距离
另外，数据保证该图联通
在以 1 为根节点的时候，如果这个根节点转移到与其相连的子节点，那么来说，和转移之前的节点相连的子树的节点的距离就要减小 1 ，假设a[t] 表示以 t 为根的子树的权值，那么来说，以之前的 1 为根的子树的权值就是a[1] , 如果是相连的子节点那么距离就要减少 a[i] , 然后对于那些非子节点的 Cost 就要增加 1,那么来说，增加的就是a[1] - a[i],所以说，总的变化就是
a[1] - a[i] - a[i]
则当前节点的Cost:
Cost[子节点] = Cost[父节点] + a[1] - a[i] - a[i]
深度优先搜索的时候遍历一下就行了，注意遍历的时候父节点子节点的关系问题
Code:
#include <bits/stdc++.h> #include <cmath> using namespace std; typedef long long ll; typedef int itn; const int maxn = 2e5 + 7; /// 链式前向星向存图 struct node { int pt;///当前节点 int net;///下一个节点的 }p[maxn << 1]; int cnt = 0; ll head[maxn]; ll pre[maxn];///预处理的权值 ll n;///总共有N个点 ll ans;///记录答案 ll a[maxn]; void _Add(int u, int v) { p[cnt].pt = v;///下一个节点； p[cnt].net = head[u]; ///cnt++; head[u] = cnt++; } /// 初始化 void _Init() { for (int i = 0; i <= n; i++) { head[i] = -1;///初始化为 -1 pre[i] = 0; } } void DFS(int x, int fa) { /// 当前节点和父节点 for (int i = head[x]; ~i; i = p[i].net) { int pt = p[i].pt;///找到对应的节点 if (pt == fa) continue;///所连向的点正好是父亲节点 DFS(pt, x);///这时候，x就成了父亲节点，pt就是子节点 /// wawawa此时的父节点是x a[x] += a[pt];///父节点的权值要加上这个子节点的权值 /// wawawa应该是+= pre[x] += (pre[pt] + a[pt]);/// 父亲节点子树的成本就是其子节点树的成本加上这个点的权 } } void dfs(int x, int fa) { if (x != 1) { ///这个点不是根节点的时候 pre[x] = pre[fa] + a[1] - 2 * a[x];///逻辑关系 } for (int i = head[x]; ~i; i = p[i].net) { int pt = p[i].pt;///获得当前指向的节 if (pt == fa) continue; dfs(pt, x);///这时候，x就成了父亲节点，pt就是子节点 } } int main() { cin >> n; _Init(); for (int i = 1; i <= n; i++) cin >> a[i]; /// 输入当前节点的权值 /// n-1条边 for (int i = 1; i < n; i++) { int u, v; cin >> u >> v; _Add(u, v); _Add(v, u);///无向图双向建边 } DFS(1, -1); dfs(1, -1); ans = 0; for (int i = 1; i <= n; i++) { ans = max(ans, pre[i]); } cout << ans << endl; return 0; } /** 8 9 4 1 7 10 1 6 5 1 2 2 3 1 4 1 5 5 6 5 7 5 8 121 **/
