【CSUST 2026】:随刄而行 树形DP

传送门

题意

分析

这种树上选点的问题，常规思路就是树上$DP$，$f[u][0/1]$表示某个点选或不选
-如果不选择$u$点，那么他对于其余每一个点的影响就是$-m$，但考虑到$m$可能为负数，所以可能会导致这个影响是正的，所以就涉及到是一次砍掉一颗子树还是一个节点一个节点去删除，所以$f[u][1]=max(-m,-m\ast sz[u])$
-如果选择$u$点，那么对于每一个子节点，选择$max(f[j][0],f[j][1])$即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n,m;
int h[N],e[M],ne[M],idx;
ll sz[N],a[N];
ll f[N][2]; //0 选  1 不选

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void pp(int u,int fa){
    sz[u] = 1;
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa) continue;
        pp(j,u);
        sz[u] += sz[j];
    }
}

void dfs(int u,int fa){
    f[u][0] = a[u],f[u][1] = max(-1ll * m,-1ll * m * sz[u]);
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa) continue;
        dfs(j,u);
        f[u][0] += max(f[j][1],f[j][0]);
    }
}

int main() {
    int T;
    read(T);
    while(T--){
        read(n),read(m);
        for(int i = 1;i <= n;i++) h[i] = -1,sz[i] = 0,f[i][1] = f[i][0] = 0,read(a[i]);
        idx = 0;
        for(int i = 1;i < n;i++){
            int x,y;
            read(x),read(y);
            add(x,y),add(y,x);
        }
        pp(1,0);
        dfs(1,0);
        dl(max(f[1][0],f[1][1]));
    }       
    return 0;
}