CodeForces 1303G-Sum of Prefix Sums (point divide and conquer + Li Chaoshu)

Title

Give a tree, each point has a bit of weight, find a path $u_1,u_2,...u_k$获 scoring number set $a_{u_1},a_{u_2},...a_{u_k}$. To maximize the prefix sum of its prefix sum, find the maximum value.

Problem solving ideas

The problem of the path on the tree will be solved by dot divide and conquer. With the idea of ​​dot divide and conquer, how to merge the current path and the existing path information.
Consider two arrays $a_1,a_2...a_{n}And{b}_1,b_2,...b_m.$Let thesumof elements be$sum$ , the sum of the prefix sum is$xsum$ . Then connect into$a_1,a_2...a_n,b_1,b_2,...b_m$之后总的$xsum=xsu{m}_a+m\ast su{m}_a+xsu{m}_b$.
This is about mmA function of$m$ . We maintainxsuma (constant) xsum_a (constant)on Li$xsu{m}_a\left(Constantnumber\right)$ and$su{m}_a\left(Ramprate\right)$ , and when the query using$su{m}_{b}Andm$ go to Li Chaoshu to querymmThe maximum value of$m$ points.
Then when querying, follow the query, and then the reverse, you can get all the information in a certain subtree as the query, and the information of other subtrees is stored as a linear function in LiChaoshucorresponding to$xsum$ .
Note that when the size of the entire tree is 2, you also need to check all other information stored in the Li Chao tree, and then the root node is used as the value of the query information.

#include<vector>
#include<iostream>
#define ll long long
#define P pair<int, int>
#define Pll pair<ll, ll>
#define pb push_back
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
const int maxn = 2e5 + 50;
struct LiChao_segment_tree{
    
    
    inline ll f(ll k, int x, ll b){
    
    
        return k*x+b;
    }
    ll k[maxn<<2], b[maxn<<2]; int lz[maxn<<2];
    int N;
    void init(int n){
    
    
        N = n; lz[1] = 1; k[1] = b[1] = 0;
    }
    inline void down(int rt){
    
    
        if(!lz[rt]) return;
        lz[rt<<1] = lz[rt<<1|1] = 1;
        k[rt<<1] = k[rt<<1|1] = b[rt<<1] = b[rt<<1|1] = 0;
        lz[rt] = 0;//AHHHHHHH!!
        return;
    }
    void update(int rt, int l, int r, ll ck, ll cb){
    
    
        if(f(ck, mid, cb) > f(k[rt], mid, b[rt])) swap(k[rt], ck), swap(b[rt], cb);
        if(l == r) return;
        down(rt);
        if(ck > k[rt]) update(rson,ck, cb);//要传的斜率较大，去右边
        else update(lson, ck, cb);//否则去左边
    }
    ll qry(int rt, int l, int r, int x){
    
    
        ll ans = f(k[rt], x, b[rt]);
        if(l == r) return ans;
        down(rt);
        if(x <= mid) ans = max(ans, qry(lson, x));
        else ans = max(ans, qry(rson, x));
        return ans;
    }
}T;
vector<int> g[maxn];
int  sz[maxn], ms[maxn], vis[maxn], totsz;
ll a[maxn];
int n, rt;
ll ans = 0;
void get_rt(int u, int fa){
    
    
    sz[u] = 1; ms[u] = 0;
    for(int v: g[u]){
    
    
        if(v == fa || vis[v]) continue;
        get_rt(v, u);
        sz[u] += sz[v];
        ms[u] = max(ms[u], sz[v]);
    }
    ms[u] = max(ms[u], totsz - sz[u]);
    if(ms[rt] > ms[u]) rt = u;
}
void init()
{
    
    
    scanf("%d", &n);
    for(int i = 1; i < n; ++i){
    
    
        int u ,v;
        scanf("%d%d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for(int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
    totsz = n;
}
void qry(int u, int fa, ll sum, ll xsum, int d){
    
    
    int f = 0; d++; sum += a[u]; xsum += sum;
    for(int v: g[u]){
    
    
        if(vis[v] || v == fa) continue;
        f = 1;
        qry(v, u, sum, xsum, d);
    }
    if(f) return;
    ans = max(ans, xsum + T.qry(1, 1, T.N, d));
}
void update(int u, int fa, ll sum, ll xsum, int d){
    
    
    d++; sum += a[u]; xsum += d*a[u];
    int f = 0;
    for(int v: g[u]){
    
    
        if(vis[v] || v == fa) continue;
        f = 1;
        update(v, u, sum, xsum, d);
    }
    if(f) return;
    T.update(1, 1, T.N, sum, xsum);
}
void cal(int u, int cursize){
    
    
    T.init(cursize);
    for(int i = 0; i < g[u].size(); ++i){
    
    
        int v = g[u][i];
        if(vis[v]) continue;
        qry(v, u, a[u], a[u], 1);
        update(v, u, 0, 0, 0);
    }
    ans = max(ans, a[u] + T.qry(1, 1, T.N, 1));
    T.init(cursize);
    for(int i = g[u].size()-1; i >= 0; --i){
    
    
        int v = g[u][i];
        if(vis[v]) continue;
        qry(v, u, a[u], a[u], 1);
        update(v, u, 0, 0, 0);
    }
}
void gao(int u, int cursize)
{
    
    
    vis[u] = 1;
    cal(u, cursize);
    for(int v: g[u]){
    
    
        if(vis[v]) continue;
        totsz = sz[v] < sz[u] ? sz[v] : cursize - sz[u];
        rt = 0;
        get_rt(v, 0);
        gao(rt, totsz);
    }
}
void sol(){
    
    

    rt = 0; ms[rt] = n+1;
    get_rt(1, 0);
    gao(rt, n);
    cout<<ans<<endl;
}
int main()
{
    
    
    //freopen("testdata.in", "r", stdin);
	init();sol();
}
/*
4
4 2
3 2
4 1
1 3 3 7
*/