Luogu P1351 联合权值 前缀和

求树上距离为二的点对权积之最值、和。

70pts：枚举每个点，bfs两层求和求最值和和。

100pts：枚举每个点，先统计所有可达节点的权值和，再枚举可达点求权值和

第二种做法较第一种的优化：

a1*a2+a1*a3+a1*a4+..a2*a1+a2*a3+...

=>a1*(sum-a1)+a2*(sum-a2)+...

好巧妙啊

要注意的一点：小心过程量爆int！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

int read(){
    int a = 0;char l = ' ',c = getchar();
    while(c < '0'||c > '9')l = c,c = getchar();
    while('0' <= c&&c <= '9')a = a*10+c-'0',c = getchar();
    if(l == '-')return -a;return a;
}

const int Maxn = 2e6+10;

struct Edge{
    int to,ne;
}edges[Maxn<<1];

int first[Maxn];
int w[Maxn];
int n,m,ans,tot,top,cnte,x;

void add_edge(int fr,int to){
    edges[++cnte] = (Edge){to,first[fr]};
    first[fr] = cnte;
}

int main(){
//    freopen("x.in","r",stdin);
//    freopen("x.out","w",stdout);
    n = read();
    for(int i = 1;i < n;i++){
        int u,v;
        u = read(),v = read();
        add_edge(u,v);
        add_edge(v,u);
    }
    for(int i = 1;i <= n;i++)w[i] = read();
    for(int i = 1;i <= n;i++){
        long long s = 0;
        int m1 = 0,m2 = 0;
        for(int j = first[i];j;j = edges[j].ne){
            int u = edges[j].to;
            s += w[u];
            if(w[u] > m1)m2 = m1,m1 = w[u];
            else m2 = max(m2,w[u]);
        }
        ans = max(ans,m1*m2);
        for(int j = first[i];j;j = edges[j].ne)
            x = w[edges[j].to],tot = (tot+x*(s-x))%10007;
    }
    cout << ans << ' ' << tot;
return 0;
}