【luoguP2986】[USACO10MAR]伟大的奶牛聚集Great Cow Gathering

题目链接

先把\(1\)作为根求每个子树的\(size\),算出把\(1\)作为集会点的代价，不难发现把集会点移动到\(u\)的儿子\(v\)上后的代价为原代价-\(v\)的\(size\)*边权+(总的\(size\)-\(v\)的\(size\))*边权

#include<iostream>
#include<cstring>
#include<cstdio>
#define int long long
using namespace std;

const int MAXN=100010;
const int MAXM=200010;

int n,a[MAXN],size[MAXN],dep[MAXN];

int Head[MAXN],num;
struct NODE{
    int to,w,nxt;
} e[MAXM];

inline void add(int x,int y,int z){
    e[++num].to=y;
    e[num].w=z;
    e[num].nxt=Head[x];
    Head[x]=num;
}

void dfs1(int x,int fa){
    size[x]=a[x];
    for(int i=Head[x];i;i=e[i].nxt){
        int v=e[i].to;
        if(v==fa) continue;
        dep[v]=dep[x]+e[i].w;
        dfs1(v,x);
        size[x]+=size[v];
    }
}

int Ans=1e18;

void dfs2(int x,int fa,int Sum){
    Ans=min(Ans,Sum);
    for(int i=Head[x];i;i=e[i].nxt){
        int v=e[i].to;
        if(v==fa) continue;
        dfs2(v,x,Sum+e[i].w*(size[1]-2*size[v]));
    }
}

signed main()
{
    scanf("%lld",&n);
    for(int i=1;i<=n;++i)
        scanf("%lld",&a[i]);
    int x,y,z;
    for(int i=1;i<n;++i){
        scanf("%lld%lld%lld",&x,&y,&z);
        add(x,y,z); add(y,x,z);
    }
    dfs1(1,0);
    int Sum=0;
    for(int i=1;i<=n;++i)
        Sum+=dep[i]*a[i];
    dfs2(1,0,Sum);
    printf("%lld\n",Ans);
    return 0;
}