[USACO10MAR]伟大的奶牛聚集

因为是英文题，题目不再重复。。

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题意

给你一棵无根树，每条边有边权，每个点有点权，要你选一个点，使每个点到这个点的距离*点权的和最小，求这个值。

思路

设dis[u]为u所有后代到它的距离*点权，sum[u]为u所有后代的点权和包括u
先以1为根Dfs一遍，预处理出所有的dis，sum
然后问题就变成了更换这棵树的根，维护dis，取所有根的dis的最小值。
所以再一次Dfs，从u转移到儿子v时，dis = dis[u] - f(边权) * sum[v] + f * (sum[u] - sum[v])

常数巨大的丑陋代码

# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# include <math.h>
using namespace std;

# define IL inline
# define RG register
# define UN unsigned
# define ll long long
# define rep(i, a, b) for(RG int i = a; i <= b; i++)
# define per(i, a, b) for(RG int i = b; i >= a; i--)
# define uev(e, u) for(RG int e = ft[u]; e != -1; e = edge[e].nt)
# define mem(a, b) memset(a, b, sizeof(a))
# define max(a, b) ((a) > (b)) ? (a) : (b)
# define min(a, b) ((a) < (b)) ? (a) : (b)

IL int Get(){
    RG char c = '!'; RG int num = 0, z = 1;
    while(c != '-' && (c > '9' || c < '0')) c = getchar();
    if(c == '-') z = -1, c = getchar();
    while(c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
    return num * z;
}

const int MAXN = 100001, INF = 2147483647;
struct Edge{
    int to, nt, f;
} edge[MAXN << 1];
int n, cnt, ft[MAXN], w[MAXN];
ll dis[MAXN], ans = 1e18, sum[MAXN];

IL void Add(RG int u, RG int v, RG int f){
    edge[cnt] = (Edge){v, ft[u], f}; ft[u] = cnt++;
}

IL void Dfs(RG int u, RG int fa){
    sum[u] = w[u];
    uev(e, u){
        RG int v = edge[e].to, f = edge[e].f;
        if(v == fa) continue;
        Dfs(v, u);
        sum[u] += sum[v];
        dis[u] += dis[v] + sum[v] * f;
    }
}

IL void Dfs2(RG int u, RG int fa, RG ll D, RG ll S){
    ans = min(ans, D);
    uev(e, u){
        RG int v = edge[e].to, f = edge[e].f;
        if(v == fa) continue;
        RG ll ss = S - sum[v], dd = D - sum[v] * f + ss * f;
        Dfs2(v, u, dd, S);
    }
}

int main(){
    mem(ft, -1);
    n = Get();
    rep(i, 1, n) w[i] = Get();
    rep(i, 1, n - 1){
        RG int u = Get(), v = Get(), f = Get();
        Add(u, v, f); Add(v, u, f);
    }
    Dfs(1, 0);
    Dfs2(1, 0, dis[1], sum[1]);
    printf("%lld
", ans);
    return 0;
}