P8269 [USACO22OPEN] Visits S

Bessie has $N$ ($2\le N\le 10^5$) cow friends (numbered $1\cdots N$), and each of them owns her own farm. For each $1\le i\le N$, friend $i$ wants to visit friend $a_i$ ($a_i\neq i$). Given a permutation $(p_1,p_2,\ldots, p_N)$ of $1\ldots N$, the visits happen as follows. For each $i$ from $1$ to $N$: - If friend $a_{p_i}$ has already left her farm, then friend $p_i$ stays at her farm. - Otherwise, friend $p_i$ leaves her farm to visit friend $a_{p_i}$'s farm. This visit produces happy moos $v_{p_i}$ times ($0\le v_{p_i}\le 10^9$). Among all possible permutations $p$, compute the maximum total number of moos that can be obtained after all visits finish.

The first line contains $N$. For each $1\le i\le N$, line $i+1$ contains two space-separated integers $a_i$ and $v_i$.

Output one integer, the required answer. **Note that the integers in this problem may require 64-bit integer types (for example, "long long" in C/C++).**

[Sample Explanation] If $p=(1,4,3,2)$, then: - Friend $1$ visits friend $2$'s farm, producing $10$ moos. - Friend $4$ sees that friend $1$ has already left her farm, so nothing happens. - Friend $3$ visits friend $4$'s farm, producing another $30$ moos. - Friend $2$ sees that friend $3$ has already left her farm, so nothing happens. So in total we get $10+30=40$ moos. On the other hand, if $p=(2,3,4,1)$, then: - Friend $2$ visits friend $3$'s farm, producing $20$ moos. - Friend $3$ visits friend $4$'s farm, producing $30$ moos. - Friend $4$ visits friend $1$'s farm, producing $40$ moos. - Friend $1$ sees that friend $2$ has already left her farm, so nothing happens. So in total we get $20+30+40=90$ moos. It can be proven that this is the maximum possible number of moos after all visits finish among all possible permutations $p$. Translated by ChatGPT 5