题目
link
题解
比较简单的做法
访问到每一个节点的时候，在儿子节点中求和，找最大值，次大值来维护答案即可
锅
1.维护sum的时候也要判断if(v == fa) continue
2.不开long long见祖宗
代码
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| #include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5 + 10;
int W[N];
int n;
int cnt = 1;
int head[N], nxt[N << 1], to[N << 1];
int maxx = 0, sum = 0;
void add(int x, int y)
{
to[++cnt] = y;
nxt[cnt] = head[x];
head[x] = cnt;
}
void dfs(int u, int fa)
{
int son_max1 = 0, son_max2 = 0;
int son_sum = 0;
for(int i = head[u];i ;i = nxt[i])
{
int v = to[i];
if(v == fa) continue;
dfs(v, u);
son_sum += W[v];
if(W[v] > son_max1)
{
son_max2 = son_max1;
son_max1 = W[v];
}
else if(W[v] > son_max2)
{
son_max2 = W[v];
}
}
maxx = max(son_max1 * son_max2, maxx);
maxx = max(maxx, W[fa] * son_max1);
for(int i = head[u]; i ; i = nxt[i])
{
if(to[i] == fa) continue;
int v = to[i];
sum = (sum + (son_sum - W[v]) * W[v]) % 10007;
}
sum = (sum + 2 * W[fa] * son_sum) % 10007;
}
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1, u, v;i <= n - 1;i++)
{
cin >> u >> v;
add(u, v);
add(v, u);
}
for(int i = 1;i <= n;i++)
cin >> W[i];
dfs(1, 0);
cout << maxx << " " << sum;
return 0;
}
|