lic lineup

1699: [Usaco2007 jan]balanced lineup queue time limit:5 Sec Memory limit:64 MB DescriptionEvery day, farmer John's N (1 Input* First line: N and Q. * 2nd. N+1: Line i+1 is the height of the first cow.* Line n+2..n+q+1: Two integers, A and B (1 Output* 1th. Q Line: All inquiries answered (the highest and lowest of the cow's height), one per line.Sample Input6 31734251 54 62 2Sample Output630HINT SourceGoldTest instructions: Interval maximum-in

g-balanced LineupTime limit:5000ms Memory limit:65536kb 64bit IO format:%i64d %i64uSubmit Status Practice POJ 3264DescriptionFor 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 shou

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

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 range of 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 T

Balanced lineup Time limit:5000 Ms Memory limit:65536 K Total submissions:34306 Accepted:16137 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 organize a game of Ultimate Frisbee with some of the cows. to keep things simple, he will take a contiguous range of cows from the milking

B. balanced lineup Time limit:5000 Ms Case time limit:5000 Ms Memory limit:65536kb 64-bit integer Io format: % LLDJava class name: Main 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 range of cows from the milking lineup to pla

Balanced lineup Time limit:5000 Ms Memory limit:65536 K Total submissions:26435 Accepted:12401 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 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

Attribute Analysis The biggest difference between Orange 2 skull King and Orange 0 is that the blood volume increases by 10000 and the attack increases by 800. Skull King of these attributes, in fact, is very poor, so do not recommend everyone to go up Orange 2. Skill Analysis "Flying Hammer" can stun targets, the damage can be neglected, the only role is to control the next. "Crit" Skull King's main output skill. "Blood sucking" team support skills, the effect is very good, this skill and

Balanced LineupTime Limit: 5000 MSMemory Limit: 65536 KTotal Submissions: 23380Accepted: 10882Case Time Limit: 2000 MSDescriptionFor 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 agame of Ultimate Frisbee with some of the cows. to keep things simple, he willtake a contiguous range of cows from the milking lineup to play the game. however, for all the cows to have fun t

At first, the face was crazy.Later I thought of maintaining a left and right hand two pointers L and R. Represents the different kinds of digital It is clear that the leftmost, legal L increases with R and does not decrease.By the way discretization, remember the number of different kinds of numbers to calculate the answer.Time complexity O (n)1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intmaxn=1e5+233;7 structzs{intV,id;} A[MAXN];8 intMP[MAXN],SM[MAXN];9 inti,j,n,m,ans,

Title: Enemy LineupStandard segment Tree Template code:#include Y the value to update **********flag inference update mode *****************************************************/void updatetree (int i, int x, int y, int flag) {int L = node[i].left; int r = node[i].right; int m = (L + r)/2; if (r = = L) {if (flag) Node[i].count + = y; else node[i].count-= y; Return } if (x Graceful line-segment tree code:#include The same is true of the two code ideas. Only the code st

appears on each feature, otherwise it will not be the same as two numbers. Because only the rightmost number increases by the same number as all the numbers on the left, they subtract the rightmost number and appear the same. */#includestring>#include#include#include#defineMoD 100007using namespacestd; intHash[mod+Ten][ the]; inta[mod][ to]; ints[mod][ to]; intK; BOOLCheckintTintXT) { inti; BOOLflag=true; for(i=0; i1; i++) if(s[xt][i]!=Hash[t][i])return false; return true; } int

Topic Portal: http://poj.org/problem?id=3264The problem is simply to find the maximum value to find the minimum value.　　#include #include#includeusing namespacestd;Const intmaxn=50010;intn,q,maxm[maxn][ -],minm[maxn][ -];voidinit () {inti,j; for(j=1;(1) for(i=1; i+ (11) {Maxm[i][j]=max (maxm[i][j-1],maxm[i+ (11))][j-1]); MINM[I][J]=min (minm[i][j-1],minm[i+ (11))][j-1]); }}intQueryintLintR) { inti,j,k=0; while(11) 1) k++; I=max (maxm[l][k],maxm[r-(11][k]); J=min (minm[l][k],minm[r-

build (Rson); - pushup (RT); About } $ - voidQueryintLintRintRtintLintR) { - if(e[rt].maxxreturn; - if(lR) { Amin=min (e[rt].minx,min); +max=Max (E[rt].maxx,max); the return; - } $ intMid= (l+r) >>1; the if(l1, l,r); the if(r>mid) query (mid+1,r,rt1|1, l,r); the //if (r the //else if (l>mid) query (rson,l,r); - //else{ in //query (lson,l,r); the //query (rson,l,r); the // } About } the the intMain () { the //fre (); +

Bare RMQ Problem1#include 2#include 3#include 4 5 using namespacestd;6 7 Const intMAXN =50010;8 9 intmin_dp[maxn][ -],max_dp[maxn][ -];Ten intMIN_MM[MAXN],MAX_MM[MAXN]; One intB[MAXN]; A intn,q; - - voidMIN_INITRMQ (intNintb[]) the { -min_mm[0] = -1; - for(intI=1; i) - { +Min_mm[i] = ((i (i-1)) ==0) ? min_mm[i-1]+1: min_mm[i-1]; -min_dp[i][0] =B[i]; + } A for(intj=1; j) at for(intI=1; i + (11) -Min_dp[i][j] = min (min_dp[i][j-1],min_dp[i+ (11))

Title Address: http://poj.org/problem?id=3264Sample Input6 31734251 54) 62 2Sample Output630Analysis: Standard template problem, can be written with line segment tree, but with rmq-st to write code relatively short.What is the difference between the maximum and minimum values for each output interval "L, R".Notice a place where the code uses the LOG2 () function, but I submit the code in C + + with the Changed to g++ submitted before AC. (Note that you do not have to use log2 (), the formula for

Title Address: POJ 3264In order to learn the online LCA algorithm, first learn RMQ ... RMQ first hair, pure template problem. Not much to say.The code is as follows:#include #include #include #include #include #include #include #include #include using namespace STD;#define LL Long Long#define PI ACOs ( -1.0)Const intMod=1e9+7;Const intinf=0x3f3f3f3f;Const Doubleeqs=1e-9;intdp1[51000][ -], dp2[51000][ -], a[51000], N;voidRMQ () {intI, J; for(i=1; i0]=dp2[i][0]=a[i]; } for(j=1;(1 for(i=1; i11; i++

Framer John had a piece of wood that he wanted to use to build a fence, and he had to ask Framer Don for help without a saw. FD request, FJ each cut open a piece of wood, will give the length of the plank of money. FJ want to spend the least, ask you for help. Idea: Definitely a replica of the merged fruit! It's OK to take the plank as a merged plank. Be careful of the range of L, N, the final ans with int is not fit, to use a long long. The code is as follows: #include #include #in

RMQ: (interval max problem)Essentially dynamic programming, with D (I, j) representing the minimum value of a section of an element that starts with a length of 2^j from I, you can use a recursive method to calculate D (i, J): D (i, j) = min{D (i, j-1), D (i + 2^ (j-1), j-1)}Because 2^j #include POJ 3264 Balanced Lineup (RMQ detailed)

; i0]=mx[i][0]=W[i]; intM= (int) (Log (n1.0)/log (2.0));//m represents the maximum value of I in the 2^i for(intI=1; i) for(intj=1; j) {Mi[j][i]=mi[j][i-1];//Mi[j][i] The minimum value can be from mi[j][i-1] and mi[j+ (1 if(J+ (11)) 11))][i-1]); Mx[j][i]=mx[j][i-1]; if(J+ (11)) 11))][i-1]); }}//the extremum of [l,r] can be calculated from [L,m],[m,r] two ranges//as long as you meet l+2^k>=r-2^k+1//2^ (k+1) >=r-l+1------>k>=log2 (r-l+1)-1;//1. When log2 (r-l+1) is an integer, K tak