题意:
思路:
有时间再补,先插眼。
时间复杂度: O ( n ) O(n) O(n)
int n, m;
int h[N], e[N], ne[N], idx;
int s[N];
multiset<int> q[N];
multiset<int> d[N];
int ans;
vector<int> tmp;
int self[N];
int jinian[N];
void add(int a, int b) {
e[idx] = b;
ne[idx] = h[a], h[a] = idx++;
}
void dfs1(int u, int fa) {
for (int i = h[u], cnt; ~i; i = ne[i]) {
int j = e[i];
if (j == fa)
continue;
dfs1(j, u);
cnt = s[j];
if (!q[j].empty())
cnt += *(--q[j].end());
q[u].insert(cnt);
if (q[u].size() > 4)
q[u].erase(q[u].begin());
}
}
void dfs2(int u, int fa) {
for (int i = h[u], cnt; ~i; i = ne[