![【题解】LuoGu6869：[COCI2019-2020#5] Putovanje](/uploads/reation/bcimg12.png)
【题解】LuoGu6869：[COCI2019-2020#5] Putovanje
原题传送门树上差分就是走n−1n-1n−1条路每次从iii走到i+1i+1i+1，所以用树上差分进行路径覆盖对于每条边可以直接买多程票，也可以买几次单程票Code：#include <bits/stdc++.h>#define maxn 200010#define LL long longusing namespace std;LL delta[maxn], odd[maxn], even[maxn], ans;struct Edge{ int to, next;}e
