题目链接

Spring tree

题目概述

给定n个铁球，重量为w_i，再给定n - 1条弹簧（可变的边权）所链接的两端，每个位置上的铁球可以相互交换。弹簧的长度为每个节点的子树边权和+1。问从根节点（1节点）开始的最大深度。

输入 #1

4
1 2 3 4
1 2
2 3
3 4

输出 #1

23

样例说明

In the test case, keep original arrangement is a good idea.
maximum depth = (1+4) + (1+4+3) + (1+4+3+2) = 23

思路

首先我们发现，更重的球会更倾向于放到深度最深的位置。

枚举最优解中深度最深的叶子 leaf，那么最优解会怎样放置呢？

首先最重的球一定给 leaf，然后考虑 leaf 的父亲 parent(leaf)，那么最重的 size(parent(leaf)) 个球一定都在 parent(leaf) 的子树中（其中 size(u) 表示 u 的子树大小）..... 依次类推。

对球的重量排序，用 w(k) 表示 k 个最重的球的总重量。

那么我们可以把节点 u 的权值设成 w(size(u)) 接着求出树上从根到树叶找出一条权值和最大的路径即可，这个可以用 O(n) 的 DP 来实现，需要注意的是根节点没有连向父亲的弹簧，所以根节点权值应该设成 0.

AcCode：

// #pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
//#include <ext/rope>
//using namespace __gnu_cxx;
using namespace std;

#define sp ' '
#define endl '\n'

template<typename T> inline T read(){
    T x = 0, f = 0; char ch = cin.get();
    while (!isdigit(ch)) f |= ch=='-', ch = cin.get();
    while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = cin.get();
    return f ? -x : x;
}
template<typename T> bool chkmin(T &a, T b){return (b < a) ? a = b, 1 : 0;}
template<typename T> bool chkmax(T &a, T b){return (b > a) ? a = b, 1 : 0;}

#define dd(x) cerr << "Line:" << __LINE__ << sp << #x << " -> " << (x) << ", "
#define de1(x) cerr << "Line:" << __LINE__ << sp << #x << " -> " << (x) << endl
#define de2(a, b, x) cerr << "Line:" << __LINE__ << sp << #a << "[" << b << "]" << "->" << x << endl

#define int ll
//#define x first
//#define y second
#define gg exit(0);
#define pb push_back
#define ls (u << 1)
#define rs (u << 1 | 1)
#define eb emplace_back
#define read() read<int>()
#define mod(x) (x + mod) % mod
#define NO cout << "NO" << endl
#define YES cout << "YES" << endl
#define all(x) (x).begin(), (x).end()
#define tmax(a,b,c) (a>b?(a>c?a:c):(b>c?b:c))
#define clar(a, b) memset((a), (b), sizeof(a))
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define fmax(a,b,c,d) (a>b?a:b)>(c>d?c:d)?(a>b?a:b):(c>d?c:d)
#define cdouble(x, a) cout << setprecision(x) << fixed << a << endl

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> PII;
typedef unsigned long long ull;

const double pi = acos(-1);
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 10;

int n, m, cnt, ans;
int tree[N], siz[N], sum[N];
vi g[N];
vi w;

void dfs1(int u, int fa){
    siz[u] = 1;
    for(auto v : g[u]){
        if(v != fa) dfs1(v, u); 
        siz[u] += siz[v];
    }
}
void dfs2(int u, int fa){
	int res = 0;
	for(auto v : g[u]){
		if(v != fa){
			dfs2(v, u);
			res = max(res, tree[v] + sum[siz[v]] + 1);
		}
	}
	tree[u] = res;
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	
    cin >> n;
    rep(i, 1, n){
        int a; cin >> a;
        w.pb(a);
    }
    sort(all(w), greater<int>());
    
    rep(i, 1, n) sum[i] = sum[i - 1] + g[i];

    rep(i, 1, n - 1){
        int u, v; cin >> u >> v;
        g[u].pb(v); g[v].pb(u);
    }
	
    dfs1(1, -1);
    
    dfs2(1, -1);

    cout << tree[1] << endl;
	return 0;
}