dispatching 101

This problem is very comfortable to do with the Chairman tree.First of all, test instructions, it is necessary to find a point I in the tree structure, and find some points in the tree to form a set, so that the sum of C in the set is not more than M, and the number of elements in the li* set and the maximumSimply think of the need to enumerate each point first, and then find the smallest k point in the subtree, so that Sigma (C[i]) (i = 1..K) does not exceed M, then l[i]*k is the optimal soluti

, time consuming" + (costtime/1000) + "seconds");} else{system.out.println (windowname+ "did not take the service task! "); Commonservice ();}} Privatevoidvipservice () {stringwindowname= "No." +windowId+ "number" +type+ "window" ; System.out.println ("Getting Tasks"); Integernumber=numbermachine.getinstance () //gets the number of the user being queued. GetVipManager (). Fetchservicenumber (); if (number!=null) {intbegintime= (int) System.currenttimemillis ();//service start time Intmaxrand=c

) Elevator failure, elevator suspension for maintenance.3, Solution(1) Single-layer docking system, some floors do not stop ( -2~7 floor)(2) Set up an infrared sensor at the door of the elevator, the elevator will stop when both the person and the button are pressed. For those who only press but no one, the elevator does not stop. (In case of entering the elevator)(3) Elevator failure, start the standby scheduling scheme. Allow other elevators to operate in a restricted mode.4, elevator scheduli

This is the prototype and framework of the graduation design, the realization of the function in the left side of the function box navigation menu.Do too bad, or the school name to the mosaic off it .... The Provincial alumni BatchLogin interface: The system three uses the ASP. NET three layer architecture, jquery Easyui front desk Show all background interaction is done with jquery Ajax. The whole page still looks very refreshing. =.=.=Key features to implement:One of the issues to be solved to

2809: [apio2012]dispatching Time Limit:10 Sec memorylimit:128 MB Description In a ninja gang, some ninjas are selected to be dispatched to the customer and rewarded for their work. In this gang, a ninja is called Master. In addition to master, each ninja has and has only one parent. To keep it private and to enhance the leadership of the Ninja, all instructions related to their work are always sent by the superior to his direct subordinates, and not b

The main idea: give a tree, each node has two values, respectively, the Ninja's salary and ninja leadership. Customer satisfaction is the point of leadership can be achieved by the number of people, provided that the sum of the salary of the person does not exceed the total amount of money.Thought: Only in the sub-tree operation, greedy think, we as long as the sub-tree cost the smallest of those points can be. So on a deep search, each to a node, put itself and all the child nodes of the balanc

[ch[x][0]]) swap (ch[x][1],ch[x][0]);h[x]=h[ch[x][1]]+1;return x;}void dfs (int x) {sum[x]=val[x],size[x]=1;for (int i=head[x];i!=-1;i=edge[i].next) {int to=edge[i].to;Dfs (to);Sum[x]+=sum[to];size[x]+=size[to];Fa[x]=merge (Fa[x],fa[to]); }While (sum[x]>m) {sum[x]-=val[fa[x]],size[x]--;Fa[x]=merge (ch[fa[x]][0],ch[fa[x]][1]); }Ans=max (Lead[x]*size[x],ans);}int Main () {H[0]=-1;scanf ("%d%d", n,m);init ();for (int i=1;i {int f,v,l;scanf ("%d%d%d", f,v,l);if (!f) {all_root=i,val[i]

];rs[y]=Rs[x]; if(posmid) build (Ls[y],ls[x],l,mid,pos); ElseBuild (rs[y],rs[x],mid+1, R,pos);}intQueryintYintXintLintRintLim) {if(L==R)returnsumv[y]-sumv[x]?lim* (S[y]-s[x])/(Sumv[y]-sumv[x]):0; intMid=l+r>>1; if(Limreturnquery (Ls[y],ls[x],l,mid,lim); returnQuery (rs[y],rs[x],mid+1, R,lim-sumv[ls[y]]+sumv[ls[x]]) +s[ls[y]]-s[ls[x]];}intMain () {n=read (); m=read (); Rep (1, N) {Addedge (read (), i); Sv[i]=read (); lv[i]=read (); } DFS (RT); Rep (1, n) Build (root[i],root[i-1],1, M,s

];Wuyi returnx; the } - WuInlinevoidsolve () { -N=getint (); m=getint (); About for(intI=1; i1; $d[0]=-1; - for(inti=n;i>=1; i--) { - while(sum[tree[i]]>m) Tree[i]=merge (L[tree[i]],r[tree[i]);//if it is out of range, delete the top of the heap (i.e. the largest weight) and merge the left and right subtree -Ans=max (ans, (LL) size[tree[i]]*able[i]); A if(i!=1) Tree[father[i]]=merge (Tree[i],tree[father[i]);//Merge this node and its father. + } the printf (Ot,ans); - }

The GestureDetector of the gesture recognition class can be used to implement double-click listening using the standard SDK class, but there is a limitation: Because GD intercepts events in the OnTouchEvent method for processing, and Android event dispatching process, in this way, if the View inside the Activity consumes click events, GD will not be able to accept click events. To this end, I have implemented a tool class that can perfectly listen to

; the } - WuyiInlinevoidAdd_edge (intXinty) { theE[++tot] =Edge (first[x], y); -FIRST[X] =tot; Wu } - AboutInlineintNew_heap (intx) { $H[++CNT_HEAP].V =x; -H[CNT_HEAP].L = H[CNT_HEAP].R = H[CNT_HEAP].DEP =0; - returncnt_heap; - } A + intMerge (intXinty) { the if(!x | |!y)returnX +y; - if(H[X].V h[y].v) $ swap (x, y); theH[X].R =Merge (H[X].R, y); the if(H[h[x].l].dep H[H[X].R].DEP) the swap (H[X].L, H[X].R); theH[X].DEP = H[H[X].R].DEP +1; - returnx; in } the theInline

); Update (x); returnx;}intInsertintW) {heap[++tot].d=0, heap[tot].sz=1; Heap[tot].lch=heap[tot].rch=0; Heap[tot].sum=heap[tot].w=W; returntot;}voidPopintx) {x=merge (Heap[x].lch,heap[x].rch);}intMain () {scanf ("%d%d",n,m); for(intI=1; i) {scanf ("%d%d%d",b[i],c[i],L[i]); Root[i]=Insert (c[i]); while(heap[root[i]].sum>m) pop (root[i]); } for(intI=n;i>0; i--) {ans=max (ans,1ll*heap[root[i]].sz*L[i]); Root[b[i]]=merge (Root[i],root[b[i]]); while(heap[root[b[i]]].sum>m) pop (Root[b[i]]);

31 3 52 2 21 2 42 3 1Sample Output6HINTIf we choose a ninja numbered 1 as a manager and dispatch a third and fourth Ninja, the sum of the salaries is 4, not exceeding the total budget of 4. Since 2 ninjas were dispatched and the manager's leadership was 3,User satisfaction is 2, is the maximum number of user satisfaction can be obtained.SolutionThis is the strategy.Solution optimality: When the whole tree is DFS, we actually enumerate each manager, take the bestand solve the problem of optimali

operate it.Summary: This pair programming is a very fresh attempt, the experience I have never had before. With very few team members, only two of them will be in better shape and in a very fast state of entry. Because my character is usually so careless, careless, so this time is very fortunate that my teammates are a careful, meticulous girl, if the full score is 100 points, I want to give LEI sister 99! (Less a point afraid of his pride Qaq) Thank you in this pair of programming to give me t

on the Elevator button, a form appears, which is a few check buttons to indicate which floors have been selected;You can use a control to represent the elevator, the position of the button control to indicate the movement of the elevator; the button control has a MouseEnter method that can be used to indicate the number of people coming in when the mouse enters the control once, the number of people is added 1;mouseleave method to show the number of people out of the elevator, That is, the numb

internal passengers on which floor, the elevator will be docked.3.2 Improvement measures: Some of the details of the problem can reduce the total operating time of the elevator to a certain extent.(1) above 12 is a foreign language or humanities-specific office teaching floor, usually take the elevator to class people will be less, so can be in a large amount of time for a person to get a lift to reach the highest floor of the Big 12 floor;(2) Find the best floor where the elevator stops when n

at any single double-decker. For peak periods, elevators can be divided into stairs and downstairs peaks. For the stairs, the elevator only accepts requests to go upstairs and does not stop at full, and once no one rides or reaches the top floor, it returns to the first floor and accepts requests downstairs. Similarly, for the downstairs peak, the elevator only accepts requests downstairs, and in full to make no stop, once no one rides or arrives at the first floor, and accepts downstairs reque

answered, cancels the request signal.The traditional elevator scheduling algorithms on the network are:1. First come first service algorithm (FCFS)2. Shortest search floor time First algorithm (SSTF)3. Scanning algorithm (scan)4.LOOK algorithm5.SAFT algorithmReal-time elevator scheduling algorithm on the network:1. First cut-off time priority scheduling algorithm2.SCAN-EDF algorithm3.PI algorithm4.fd-scan algorithmElevator Group control scheduling on the network1. Give expert system a bit of gr

Specifications Stop Layer Actual speed Flat-layer Time Average Transit time A1 20 Floor 20 Stations -2,-1,1,9,11,13,15,17 About 0.8M/S 8.7s 37.2s A2 18 Floor 18 Stations 1,9,11,13,15,17 About 0.8M/S 8.7s 36.5s B1 20 Floor 20 Stations -2,-1,1,8,10,12,14,16,18 About 0.8M/S 8.7s 42.2s B2 18 Floor 18 Stations 1,8,10,12,14,16,18 Ab

need more external requests to stop. The elevator scheduler does not know how many passengers are waiting for each layer, and it does not know how many passengers will appear. This is the same as the real-world situation.At the peak of each class and class, about---18 people.To solve the problem: avoid the "bus" worst case, the same floor has several elevators can reach, the floor has a user call, which elevator to respond to the call, should be running to pass through the layer, or the nearest