12月USACO競賽黃金組別真題解析~USACO競賽沖刺班招生中

Bessie recently discovered that her favorite pop artist, Elsie Swift, is performing in her new Eras Tour! Unfortunately, tickets are selling out fast, so Bessie is thinking of flying to another city to attend the concert. The Eras tour is happening in N(2≤N≤750) cities labelled 1…N, and for each pair of cities (i,j) with i<j there either exists a single direct flight from i to j or not.

A flight route from city a to city b (a<b) is a sequence of k≥2 cities a=c1<c2<?<ck=b such that for each 1≤i<k, there is a direct flight from city ci to city ci+1. For every pair of cities (i,j) with i<j, you are given the parity of the number of flight routes between them (0 for even, 1 for odd).

While planning her travel itinerary, Bessie got distracted and now wants to know how many pairs of cities have direct flights between them. It can be shown that the answer is uniquely determined.

考情分析

從相鄰位置來考慮比較簡單，如果i到i+1路徑為奇數(shù)，說明i到i+1有邊，反之沒有。

考慮任意點(diǎn)i和j的時候，只需要利用區(qū)間dp的思想計算出i到j(luò)除了直接到達(dá)的路徑有多少條，看一下是否滿足奇偶性即可。

時間復(fù)雜度:   O(n^3)

知識點(diǎn)：區(qū)間dp

Bessie is going on a trip in Cowland, which has N (2≤N≤2⋅10^5) towns numbered from 1 to N and M (1≤M≤4⋅10^5) one-way roads. The i-th road runs from town ai to town bi and has label li 

(1≤ai,bi≤N, 1≤li≤10^9).

A trip of length k starting at town x0 is a sequence of towns x0,x1,…,xk, such that there is a road from town xi to town xi+1 for all 0≤i<k. It is guaranteed that there are no trips of infinite length in Cowland, and that no two roads connect the same pair of towns.

For each town, Bessie wants to know the longest possible trip starting at it. For some starting towns, there are multiple longest trips - out of these, she prefers the trip with the lexicographically minimum sequence of road labels. A sequence is lexicographically smaller than another sequence of the same length if, at the first position in which they differ, the first sequence has a smaller element than the second sequence.

Output the length and sum of road labels of Bessie's preferred trip starting at each town.

考情分析

首先題目保證給出圖是DAG，那么我們要求最長路的話就可以利用拓?fù)渑判?bfs來解決。

難點(diǎn)在于題目要求字典序最小，考慮如何快速比較兩條出邊的字典序大小

由于題目權(quán)值可以相同暴力走下去比較時間復(fù)雜度會被卡到$O(n^2)$

為了優(yōu)化比較過程，我們希望能夠快速得到兩個最長路相同的點(diǎn)他們之間的字典序關(guān)系。

所以我們可以動態(tài)地對最長路相同的點(diǎn)的內(nèi)部做一個排序，比較的時候就只需要拿出之前維護(hù)好的排序結(jié)果去做判斷，判斷時間復(fù)雜度就可以做到O(1)

總時間復(fù)雜度：O(nlogn)

知識點(diǎn)：拓?fù)渑判颍值湫?/span>

Farmer John is distributing haybales across the farm!

Farmer John's farm has N (1≤N≤2⋅10^5) barns, located at integer points x1,…,xN (0≤xi≤10^6) on the number line. Farmer John's plan is to first have N shipments of haybales delivered to some integer point y (0≤y≤10^6) and then distribute one shipment to each barn.

Unfortunately, Farmer John's distribution service is very wasteful. In particular, for some ai and bi

(1≤ai,bi≤10^6), ai haybales are wasted per unit of distance left each shipment is transported, and bi haybales are wasted per unit of distance right each shipment is transported. Formally, for a shipment being transported from point y to a barn at point x, the number of haybales wasted is given by

Given Q (1≤Q≤2⋅10^5) independent queries each consisting of possible values of (ai,bi), please help Farmer John determine the fewest amount of haybales that will be wasted if he chooses y optimally.

考情分析

觀察1：查詢的次數(shù)很多，可以考慮預(yù)處理，由于不需要頻繁修改區(qū)間，使用前綴和即可。

觀察2：題目保證數(shù)軸x1...xN上存在一個極小值點(diǎn)y，通過等式變形/反證法可以證明y必定為x1...xN中的一點(diǎn)，同時路徑總和f(y)是一個兩邊往中間遞減的單峰函數(shù)，可以使用三分法快速逼近y的最優(yōu)值，三分法模板。

時間復(fù)雜度: O(Tlogn)

知識點(diǎn)：三分