large family tree template

You can persist line tree template problems.#include #include#include#include#include#include#include#includeusing namespacestd;intn,q,tot,a[110000];introot[2100000],left[2100000],right[2100000];intval[2100000];voidInsert (Const intLConst intRConst introot_l,int Root_r,Const intd) {Val[root_r=++tot]=val[root_l]+1; if(L==R)return ; intmid=l+ (r-l) >>1); if(dmid) {Right[root_r]=right[root_l]; Insert (L,MID

exceeding 109 by their absolute values---the array for which th E answers should be given.The following m lines contain question descriptions, each description consists of three numbers:i, J, and K (1 OutputFor each question output of the answer to it---the k-th number in sorted A[I...J] segment.Sample Input7 31 5 2 6 3 7 42 5 34 4 11 7 3Sample Output563HintThis problem have huge input,so please use C-style input (scanf,printf), or your may got time limit exceed.algorithm and topic Analysis: gi

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Modern people, especially the middle-aged generation of pressure is too large-not only have a tense work and complex interpersonal relationships, but also need to solve a lot of practical problems: the pension problem, education of children, rising prices, inflation, currency depreciation problems, employment pressure increases ... Therefore, if the use of the first use of the property and make it to preserve or even appreciate, become more "" "is a l

family, and then connect the next array along the way. The next array is fantastic. Every moment, it actually forms a number of two-way linked lists hanging together to represent the way of a flower inside.－－－－/\| | || |V V－－－－－－－－－－／　　　　　　　　　　＼+---+| || |+---->s How to do the maximum match of the power graph.Read the paper ... Use a method like km, however, to give each flower another weight. It's really complicated ...There's a guy who wrote the c

, r,dep+1);}////query interval k large value, [L,r] is a large interval, [l,r] is a query of the inter-communityintQueryintLintRintLintRintDepintK) {if(L==R)returnTREE[DEP][L];intMid= (L+R)/2;/// by the number of elements of the difference, determine whether the left or right interval to find intcnt=toleft[dep][r]-toleft[dep][l-1];/// interval [l,r], the number of elements assigned to the left interval

in the subtree with this node root are words prefixed with "AP"! And all words prefixed with "AP" are in a subtree rooted in this node ~ "Little ho surprises.""Yes, and you have an idea of how to solve my problem?" "Little Hi Chase asked.""Hmm ... That is, every time I get your string, I find the corresponding node in the tree, and then count how many of the nodes are there? "Little Ho is not very sure to answer:" But this ... It seems that in the wo

!!Attention:Array of functional segment trees to open up a bit!!--------------------------------------------------------------------------------------------------------------- ------This problem is the template problem, first discrete, to find the interval K big time lx=root[l-1],rx=root[r], both sides at the same time to continue to make poor, look at the number of left children, if K is more large to redu

Basic implementation of the trie tree The insert, delete, and find operations of the letter tree are simple. Use a repeat loop, that is, find the child tree corresponding to the first letter in the I cycle, and then perform the corresponding operation. To implement this letter tree, we can save it with the most common

; The end of the edge int weight;//edge weight }edata; Graph is the corresponding structure of the adjacency matrix.Vexs is used to save vertices, Vexnum is the number of vertices, edgnum is the number of edges, and the matrix is a two-dimensional array to hold the matrix information. For example, matrix[i][j]=1 means "vertex I (i.e. vexs[i])" and "Vertex J (i.e. Vexs[j])" are adjacency points; matrix[i][j]=0, they are not adjacency points.Edata are the structures that correspond to the e

Introduction: http://blog.csdn.net/metalseed/article/details/8045038 Comprehension (drawing more vivid): http://blog.csdn.net/regina8023/article/details/41910615 Chairman Tree three frequently used points: 1. Enquiry Interval K-Large (online) 2. query interval k large and update node operation (offline, offline mainly to the line segment

root node, the maximum length is the diameter of the tree, this value itself is relatively large, and the complexity of searching for any interval on the online segment tree is also log2 (N). In this way, the time complexity of searching once is higher than that of direct brute force, obviously, this is not the case. We have to figure out a solution. Can we sear

], [L, R], [r+1, +]1, K[1,l-1]Obviously x.sum (k) = = 0 and y.sum (k) = = 0-ans = x.sum (k) *k + y.sum (k) = 0*i+0 = 0 The result is in accordance with the actual.2, K[l, R]X.sum (k) * k = x[l] * k = val * k, y.sum (k) = y[l] = 1 * val * (L-1)Ans = val * k-val * (L-1) = Val * (k-(L-1));3, K[r+1,]X.sum (k) * k = (X[l] + x[r]) * k = 0 * k = 0;Y.sum (k) = Y[l] + y[r] =-val * (L-1) + val * R = val * (r-l+1) = val * sizeX.sum (k) * k + y.sum (k) = val * sizeProof is complete.The two

Test instructions: Give you a number, each time to give you an interval, to find the number of K-Large section#include using namespacestd;Const intmaxn=100010;inttree[ -][MAXN];//value that represents each position in each layerintSORTED[MAXN];//the number that has been sorted wellinttoleft[ -][MAXN];//Toleft[p][i] Indicates that layer I has several points from 1 to I into the leftvoidBuildintLintRintDEP) { if(L==R)return; intMid= (l+r) >>1; intsam

Use this template can direct a off HDU 2665 Kth number This problem!/* Chairman tree for interval k large Template: * Template Special Description: * Each tree is maintained from 1 to CNT subscript information */#include Copyrigh

#include#include#include#include#include#define EPS1e-8UsingNamespace Std;/*Minimum tree graph template-Zhu Liu algorithmTemplate Description: Point marking must be 0-(N-1)Must be removed to its own point (the Benquan of its own side is infinitely large)*/#define M109#define TypeDoubleConstTypeInf=1e20;struct Node{int UV;TypeCost;}e[M*m+5];int Pre[M],id[M],Vis[M]

The main topic: for a sequence, each query interval [l,r] k tree.Analysis:Chairman Tree Template title ProgramKthtree;type Point=RecordL,r,s:longint; End;varT:Array[0..100000* -] ofPoint ; A,b,id,root:Array[0..100000] ofLongint; N,i,m,x,y,k,v,len:longint;procedureqsort (l,h:longint);varI,j,t,m:longint;beginI:=l; j:=h; M:=a[(I+J)Div 2]; Repeat whileA[i] DoInc (i); whileM DoDec (j);ifI Then beginT:=a[i]; A[I

Can and HeapThe consolidated heap can be supported./*Dagen*/structheap{intL,r,w;} H[n];intRt[n];//Subscript of the root of the I-heap/*merging a heap with X, y as root*/inlineintMergeintXinty) { intT; //One of the heaps is empty if(!x| |! Yreturnx+y; //causes x, y two root to be large if(h[x].wH[Y].W) { intt=x;x=y;y=T; } //keep the balance on both sides of the heapH[x].r=merge (Y,H[X].R); T=h[x].l;h[x].l=h[x].r;h[x].r=T; returnx

currently do not need to tube if(Q[P1]//For the left side of the interval statistic modified value (left means that the operation is left so that the query on the right will have an impact) if(!q[p1].id) Sum+=q[p1].val; Tmp[t++]=q[p1++]; }Else{//For the right interval update query if(q[p2].id==1) Ans[q[p2].val]-=sum;Else if(q[p2].id==2) Ans[q[p2].val]+=sum; Tmp[t++]=q[p2++]; } } while(p1//No query anymore while(P2if(q

P3372 [TEMPLATE] Line Segment tree 1 query and interval modification, p3372 Line SegmentDescription For example, if you know a sequence, you need to perform the following two operations: 1. Add x to each number in a certain range 2. Obtain the sum of each number in a certain range.Input/Output Format Input Format: The first line contains two integers N and M, indicating the number of numbers in the series