题目大意

给你一棵树，距离为2的两个点代价为$w_i\ast w_j$，问你最小代价和代价之和

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解题思路

搜索这棵树，每次拿父亲和子节点一起计算即可

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代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define wyc 10007
#define N 200100
using namespace std;
int n, x, y, tot, anssum, ansmax, s[N], head[N];
struct rec
{int to, next;
}a[N<<1];
void add(int x, int y)
{a[++tot].to = y;a[tot].next = head[x];head[x] = tot;
}
void dfs(int x, int fa)
{int sum = s[fa] % wyc, maxx = s[fa];for (int i = head[x]; i; i = a[i].next)if (a[i].to != fa){anssum = (anssum + sum * s[a[i].to] % wyc) % wyc;//求代价之和ansmax = max(ansmax, maxx * s[a[i].to]);//求最大值dfs(a[i].to, x);sum = (sum + s[a[i].to]) % wyc;maxx = max(maxx, s[a[i].to]);}return;
}
int main()
{scanf("%d", &n);for (int i = 1; i < n; ++i){scanf("%d%d", &x, &y);add(x, y);add(y, x);}for (int i = 1; i <= n; ++i)scanf("%d", &s[i]);dfs(1, 0);printf("%d %d", ansmax, anssum * 2 % wyc);return 0;
}