lic lineup

This question is very basic, almost a StAlgorithm. This question can be used to practice the st Algorithm for the first time. Balanced lineup Time limit:5000 Ms Memory limit:65536 K Total submissions:23289 Accepted:10827 Case time limit:2000 ms Description For the daily milking, Farmer John'sNCows (1 ≤N≤ 50,000) always line up in the same order. One day farmer John decides to orga

DescriptionFor the daily milking, Farmer John's n cows (1≤ n ≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has fun they should not differ too much in height.Farmer John has made a list of Q (1≤ q ≤200,000) Potential groups of cows and t

DescriptionFor the daily milking, Farmer John's n cows (1≤ n ≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has fun they should not differ too much in height.Farmer John has made a list of Q (1≤ q ≤200,000) Potential groups of cows and t

DescriptionFor The daily milking, Farmer John's n cows (1≤ n ≤50,000) always Line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has fun they should not differ too much in height.Farmer John has made a list of Q (1≤ q ≤200,000) Potential groups of cows

Reprint Please specify source: http://www.cnblogs.com/StartoverX/p/4618041.htmlTopic:Balanced lineuptime limit:5000ms Memory limit:65536ktotal submissions:38271 accepted:17936case time Limi T:2000msdescriptionfor The daily milking, Farmer John ' s N cows (1≤n≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking

The Cow lineup Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5367 Accepted: 3196 DescriptionFarmer John ' s n cows (1 1 5 3 2 5 1 3 4 4 2 5 1 2 3Farmer John's acute mathematical mind notices all sorts of the properties of number sequences like that above. For instance, he notices and the sequence 3 4 1 3 is a subsequ

P2880 [usaco 07jan] Balanced Lineup and p2880usaco 07janBackground Description: Every day, John's N (1 Input: Row 3: N, Q Rows from 2nd to N + 1: height of each ox Line N + 2 to N + Q + 1: two integers A and B, representing all the cows from A to B. (1 Output: Output one number per line, which is the difference between the maximum number and the minimum number.Description For the daily milking, Farmer John's N cows (1 ≤ N ≤50,000) always line up in

Vehicle and channel equipment selectionEquipment: spell strength> life pack> dual protection> cool-down and reduced. The vehicle channel is a spell-type meat shield with certain output capabilities.Che Qu hero combinationShaoguan lineup 1: che qu + Bai Xiang Jing + tie fan PrincessOf course, this combination is not used to fight difficult BOSS, but to fight the 50-second customs clearance of the little monster level, the car Channel, white elephant, a

Lineup matching and heroic status: 3 stars of the ancient giants (in addition to the Necklace 5 purple +1, three-level intelligence, endurance, French resistance, plus a random three-level gem, skill full) 3 Star Fish captain (all Violet, Level 4 strength, Agile gem, 3 attack, French anti-gem, full skill) 4 Star Double Head dog (All Violet, Level 3 intelligence, attack, plus 2 random 3 gems, full skill) 3 Star Milk (4 pieces of violet +1,3 intelli

　　hit Kill Blood Count lineup recommended: Tides + Death Prophet + bounty hunter + Warlock + machine Armour Raccoon 　　Kill blood Count Line of thought: Before a player blood count using a double front row of play, such as the tide, so that can be very good use of skill effect, reduce the count control role, and the Phantom Thorn, such as jumping out to hurt the hero, it is easy to be charged, so it is inappropriate to play. But the bounty hunter i

Recommended pingshanwwill combinationDizzy gun * 3 + Leijun * 2 + Ice method * 2 + crossbows * 1Pingshan lineup layout map:In the above lineup, the reason why we put the stun gun in front is to slow down the speed of the enemy. Here, we should pay attention to the use of the wrong gun to make the most out of the box, then through ice and crossbows, the enemy can be weakened. Here, the crossbows can also be

Balanced LineupTime limit:5000ms Memory limit:65536kTotal submissions:36813 accepted:17237Case Time Limit:2000msDescriptionFor the daily milking, Farmer John's N cows (1≤n≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has fun they should

Title index: http://poj.org/problem?id=3264 Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 58842 Accepted: 27537 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1≤n≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take

Topic Link: Click to enterIn fact, this dynamic query interval maximum minimum value of the problem, the solution is a lot, like a line tree and a tree array can be done. The efficiency of the ST algorithm is the same as the above two, but the coding is much simpler.The ST algorithm is an algorithm that uses recursive thought to calculate, so that DP (I,J) represents the minimum value from a section of an element that has a length of 2^j from I, then DP (I,J) =min (DP (I,J-1), DP (i+2^ (J-1), j-

Description For the daily milking, Farmer John'sNCows (1 ≤N≤ 50,000) always line up in the same order. One day farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous rangeOf cows from the milking lineup to play the game. However, for all the cows to have fun they shocould not differ too much in height. Farmer John has made a listQ(1 ≤Q≤ 200,000) potential groups of cows and Th

; if(RreturnQuery1 (L, R, Lson); Else if(L>mid)returnQuery1 (L, R, Rson); Else { intLMAXS =Query1 (L, Mid, Lson); intRmaxs = Query1 (mid+1, R, Rson); intMaxs =Max (LMAXS, RMAXS); returnMaxs; }}intQuery2 (intLintRintRT) { if(Tree[rt].l==l tree[rt].r==R)returnTree[rt].mins; intMid = (TREE[RT].L+TREE[RT].R)/2; if(RreturnQuery2 (L, R, Lson); Else if(L>mid)returnQuery2 (L, R, Rson); Else { intLmins =Query2 (L, Mid, Lson); intRmins = Query2 (mid+1, R, Rson); intmins =min (lmins

The question is: a farmer has a nheaded ox, and the height of each ox is different, so that you can find out the height difference between the highest ox and the lowest ox in the specified range. Solution: When I saw the question, I thought of the Line Segment tree. Then I used the recursive line segment tree. At the beginning, I had to replace it with a non-recursive version. Then I optimized the input and output for, this made me feel the beauty of optimization !!! 1 #include For non-recursiv

Uses a one-dimensional array. (Because the one-dimensional number is a Complete Binary Tree, the time is slower than using the child node pointer to build the tree, but you can basically ignore =-=) #include Poj3264balanced lineup (Basic Line Segment tree)

Yesterday afternoon thought for a long time did not think out, sure enough is very weak, thinking ability is low.Like the idea of a monotone queue, maintenance of a length of the $k+1$ slider, each time the statistics $ans$ can be#include No more."Bzoj 3048" "USACO2013 Jan" Cow lineup Slider thought

is obviously a rmq problem, multiply the method, maintain the area of the maximum/small value on the line.1 varN,i,j,q,f,t,times:longint;2hmin,hmax:array[0.. -,0..50000] of Longint;3 function min (a,b:longint): Longint;4 begin5 ifA>B then exit (b)6 Elseexit (a);7 end;8 function Max (a,b:longint): Longint;9 beginTen ifA>B then exit (a) One Elseexit (b); A end; - begin - readln (n,q); the fori:=1to n Do - begin -READLN (hmax[0, I]); -hmin[0, i]:=hmax[0, I]; + end; - fori:=1to Tru