dispatching 101

GeoServer learning note (7): Servlet and HTTP dispatching process 4 Su Weimin http://www.gisdev.cn/http://blog.csdn.net/suen/ Date: All copyrights reserved. If you need to reprint the information, contact the author and indicate the source in a conspicuous position. The HTTP dispatch process (http://blog.csdn.net/suen/archive/2009/11/02/4759332.aspx) before GeoServer version 1.6 was introduced before, and now the dispatch process after version 1.6 is

); ElseLink (t[u].rc,mid+1, r,p,c);}voidMerge (intAMP;U1,intU2) { if(!u1| |! U2) {u1=u1+u2;return ;} T[U1].C+=t[u2].c;t[u1].sum+=t[u2].sum; Merge (T[U1].LC,T[U2].LC); Merge (t[u1].rc,t[u2].rc);} ll c[110000],d[110000];ll Query (intU1,intU2,intLintr,ll s) { if(L==R)returnMin (t[u2].c-t[u1].c,s/l); ll CC=t[t[u2].lc].sum-t[t[u1].lc].sum; ll K=t[t[u2].lc].c-t[t[u1].lc].c; intMid= (L+R)/2; if(Cc>s)returnquery (t[u1].lc,t[u2].lc,l,mid,s); Else returnK+query (t[u1].rc,t[u2].rc,mid+1, r,s-cc);}

https://www.lydsy.com/JudgeOnline/problem.php?id=2809The board question WA a bit because output ans has no LLD1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 Const intmaxn=100100;9 intn,m;Ten intch[maxn][2]={},siz[maxn]={},sum[maxn]={},cnt[maxn]={},rt[maxn]={}; One intfa[maxn]={},val[maxn]={},l[maxn]={}; A inty[maxn],nex[maxn]={},head[maxn]={},tot=0; - Long Longans=0; - voidInitintXintYi) { they[++tot]=yi;nex[tot]=head[x];head[x]=tot; - } - voidUpdata (intx) {

; Getpeople (Rt[r[i]],rt[l[i]-1],1, N); Mmax=max (mmax,ld[i]*P); } printf ("%lld\n", Mmax); return 0;}Left-leaning tree:#include #include#include#includeusing namespaceStd;typedefLong LongLL;intn,m;structnode{intX,y,next;} a[1100000];intlen,last[1100000];structheap{intL,r; LL c,d; Heap () {L=r=d=0; }}h[1100000];introot[1100000];voidInsintXinty) {Len++; a[len].x=x;a[len].y=y; A[len].next=last[x];last[x]=Len;}intMarge (intXinty) { if(x==0|| y==0)returnx+y; if(H[X].C//maintenance of large pil

Portal: http://www.lydsy.com/JudgeOnline/problem.php?id=2809Save the code that can be stacked with the template.#include 　　bzoj2806 [apio2012]dispatching "can be stacked"

Left-leaning tree.Each child node maintains a large heap, traversing a son to merge himself, and merging to find that the money is not enough to delete the top of the team.//Achen#include #include#include#include#include#include#include#include#includeConst intn=100007; typedefLong LongLL;using namespacestd;intn,root,rt[n],ch[n][2],sz[n],f[n]; LL m,v[n],cs[n],w[n],ans;templatevoidRead (T x) {CharCh=getchar (); x=0; T f=1; while(ch!='-' (ch'0'|| Ch>'9')) ch=GetChar (); if(ch=='-') f=-1, ch=GetCh

Welcome to the original source-- Blog Park-zhouzhendong go to the blog Park to see the puzzleTopic Portal-BZOJ2809Test Instructions SummaryN points make up a tree, each point has a leadership and cost, you can make a point when the leader, and then in this point of the sub-tree select some cost and not more than M points, get leadership leadership by the number of points selected (leader can not be selected) profit. The maximum value of the profit. n≤100000 SolvingDo a tree-like DP operation.Ma

the budget. Sample Input5 40 3 31 3 52 2 21 2 42 3 1Sample Output6Idea {Left side of the tree understand, but through the problem can still be found not skilled.First, it is easy to think of a tree DP approach to merge the solutions upward.It is thought that the use can be stacked. But is it a Dagen or a little Gan?Joscheggen ( which I wrote most initially ), should be counted sequentially, until and above m , all other popsSee, this complexity is relatively high. ( But I don't have TLE,WA ..

By the members of ntdev and ntfsd | published: 01-may-03 | modified: 01-may-03 IRP dispatching and handling If a dispatch routine returns status_pending, the IRP passed intoDispatch routine must be marked pending with iomarkirppending (). If an IRP is marked pending with iomarkirppending () in a dispatch routine,That dispatch routine must return status_pending. If an IRP is to be marked pending, iomarkirppending must be called beforeThe IRP is ac

voidInit () {ess=0; Memset (G,0,sizeof(g));} - voidUpdateintx) {sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1; sm[x]=sm[ch[x][0]]+sm[ch[x][1]]+v[x];} - intMergeintXinty) { the if(!x| |! YreturnX+y;if(V[x]1]=merge (ch[x][1],y); -Swap (ch[x][0],ch[x][1]); Update (x);returnx; - } - voidPopintx) { + intY=merge (ch[x][0],ch[x][1]); ch[x][0]=ch[x][1]=0; sz[x]=1; SM[X]=V[X]; x=y; - } + voidDfsintx) { ARt[x]=x; for(intI=G[X];I;I=ES[I].N) DFS (es[i].t), rt[x]=merge (rt[x],rt[es[i].t]); at while(sz[rt[x

We need to enumerate the root and then choose as many points from its subtree as possible and pay no more than M, but the complexity of the violence is not right.Then consider merging the tree from bottom up (there is only one node in each tree, which is itself)Each tree is a heap, we maintain the number of nodes in the tree and the sum of the salaries, and when combined, the top salary of the heap is constantly popped up until the sum of the salary does not exceed m, then the answer is updated

of C discretization after the block (note is the weight C instead of the sequence s), for the non-modified interval query with the team to transfer, the weight of the block is O (1), the query is O (sqrt (n)), the complexity is still O (n*sqrt (n)).The next step is the detail question. We first DFS to find the DFS sequence and after the interval sort (with the left end of the block as the first keyword, the right end as the second keyword), and then the tree nodes according to the size of C, th

(T[X].W//Dagen theT[x].r=mge (t[x].r,y); + if(dep[t[x].l]DEP[T[X].R]) swap (T[X].L,T[X].R); -dep[x]=dep[t[x].r]+1; $ returnx; $ } - voidDelintAMP;X) {// -x=mge (T[X].L,T[X].R); the //Fa[fa[x]]=fa[x]; - return;Wuyi } theInline LL Top (intx) { - //int X=find (x); Wu returnT[X].W; - } About // $LL ans=0; - voidDFS (intu) { -rt[u]=++CNT; -t[cnt].w=a[u].c; Asum[u]=a[u].c; +size[u]=1; the for(intI=hd[u];i;i=e[i].nxt) { - intv=e[i].v; $ DFS (v); thesum[u]+=Sum[v]; thesize[u

Title: Given a tree, select some points in a subtrees tree, salary and not exceed m, the number of points * The maximum of leadership capacity of a sub-root nodeConsider for each node, we maintain a data structure in which greed seeks employment with small salaries.Every single node violence reconstruction must not be possible. We consider the available data structures. Each node merges the information of the child nodes directly toA treap that can be combined with a revelation. can also be used

Title: Given a tree, select some points in a subtrees tree, salary and not exceed m, the number of points * The maximum of leadership capacity of a sub-root nodeConsider for each node, we maintain a data structure in which greed looks for small payroll hires.Each node violence reconstruction must not be, we consider the data structure, each node will directly merge the information of the child nodeTreap can be combined with heuristics, or can be used with a heapToday deliberately to learn this p

. In the future programming we will continue to improve our normative writing and code algorithm effectiveness and so on. Experience and summary of programming process Through two weeks of time, although the final rough finished the knot on the work, but the harvest is heavy, this programming brings me a kind of different experience, let me feel the importance of cooperation, in programming there are many times our opinion is not agreed to add each person's code style and norms are

handover work is not know what the variable represents the meaning, but the mouth to their own requirements are particularly strict, Always make the code as clear as possible. This is the place where I need to learn. At the beginning of the code test I was not too familiar with the help and encouragement of the mouth to gradually adapt to the job. Initial completion of work: Due to time is more tense, our task can only be said to be preliminary completed, and did not take into account the e

throughout the process, our procedures are improved step-by-step, until the final molding, without the exchange of discussions between each other, but also to both sides have learned a lot of new knowledge. Our cooperation in this pair of programming projects is very enjoyable. Although the final finished product does not fully meet all requirements, but this project is important to let a beginner Java I consolidated a lot of basic knowledge and also learned a lot of new knowledge, I believe th

For this pair of tasks, because themselves are not in school, so there is no good way to communicate with my teammates, so my task is to establish the implementation method, and proposed algorithm, here very grateful to my two teammates, they are very responsible, as a team, very good to complete the task here, To exert their own power to the maximum.The feeling of this task is that they used to be a person to do a program Ah, task Ah, very few and one or a few people to carry out such cooperati

Material Dispatchearthquake in a certain area, the disaster area has been very difficult, the victims urgently need some tents, clothing, food and plasma and other supplies. The road to the disaster area is full of landslides,70% above the road damage, the bridge is all destroyed. The state immediately launched emergency preparedness plans, the history of the largest non-combat air transport operations, ready to drop urgently needed supplies to the disaster area. one side has difficulty , p plus