acm queue

260 001ENQUEUE 259002ENQUEUE 259003ENQUEUE 259004ENQUEUE 259005DEQUEUEDEQUEUEENQUEUE 260002ENQUEUE 260003dequeuedequeuedequeuedequeuestop0Sample OutputScenario #1101102103201202203Scenario #2259001259002259003259004259005260001According to the title, this team queue, only need to consider whether there are people in the queue in the current team,Therefore, each team can send a representative to

more than 3.Here the use of two queue alternating, for 12 off the case, only 1 of the report will be in the other team.For 123, it is natural to report 12 to another team.Code:#include #include#include#include#include#includeSet>#include#include#include#includestring>#include#defineLL Long Longusing namespacestd;intN;voidWork () {Queueint> q[2]; BOOLnow =0; intnum; for(inti =1; I i) Q[now].push (i); for (;;) { if(Q[now].size () 3)

; + } - }task[max]; the * /*find the highest priority between parent and child in place of parent*/ $ intReplacepa (intx)Panax Notoginseng { - intPA =x; the if(Rchild (x) N) +PA =PRIOR (Task, X, PRIOR (Task, Lchild (x), Rchild (x))); A Else if(Lchild (x) N) thePA =PRIOR (Task, X, Lchild (x)); + returnPA; - } $ $ /*Filter Down (adjust the heap downward)*/ - voidPercolatedown (intx) - { the intRP =Replacepa (x); - while(RP! =x) {Wuyi swap (TASK[RP], task[x]); thex =R

this stack structure, can not directly out of the stack 3 or 4, must first out of the Stack 5 can!!!So the auxiliary stack with the data stack at the same time out of the stack or into the stack without a direct ordering to save all the data, you can follow the following ideas: (see from here)Expand:Implement a queue, with out of the team (DeQueue), queue (EnQueue), take the minimum element (Getmin) three

; -queueint>Q; - Q.push (n); - while( !q.empty ()) + { -Cur =Q.front (); + Q.pop (); AV[cur] =1; at if(cur = =k) - { -coutEndl; - return; - } - in for(intI=1; i3; i++) - { to if(i = =1) +temp = cur+1; - Else if(i = =2) thetemp = cur-1; * Else if(i = =3) $temp = cur*2;Panax Notoginseng //determine the range of temp first, lest v[] array out of bounds, ps:temp cannot be equal to MAXN

Wa is a dead question. A person goes from the starting point to the target point. There is a camera in this process. The camera's illumination range is two units of length, including the camera's own position. There are two options to avoid being exposed. Wait 1 s in one location, or sit in the box (3 S) and take one step to spend 1 s. The camera turns clockwise every second. 1.4s has a loop, so vis [r] [C] [SEC] is the optimal solution in four situations. 2. No need to display the diagram. Eac

Analysis: Can be done with line segment tree, but feel trouble. With a priority queue, the first K-large number is retained each time you insert.#include HDU ACM 4006 The kth great number segment tree? Priority queue?

admission +1 * if( This-GT;NC = =NULL) $ This-GT;NC =NewNode (nc,null);Panax Notoginseng Else{ -Node *node =NewNode (NC, This-NC); the This-GT;NC =node; + } A return; the } + - /*Topological sorting*/ $ voidtopology () $ { - for(inti =1; I ) - if(!mark[i]) topology[++lt] = i;//City with an entry level of 0 the //Main Content - for(inti =1; topology[i];i++)Wuyi { the intcur = topology[i];//the City-number - //Traverse a

]);}voidSolveadd (intFintS//+{Node temp; Fat[s]=F; Val[s]=val[f]+1; Temp.key=s; Temp.value=Val[s]; Ans.push (temp); return;}voidSolvesub (intF//-{Node temp; ints; FAT[F]=0; while(!Ans.empty ()) {Temp=Ans.top (); if(Find (temp.key) = =1) return; S=Temp.key; while(fat[s]!=0)//The chain is assigned 0.{Fat[s]=0; } ans.pop (); } return;}intMain () {intt,n,m,coun;scanf ("%d",t); node temp; while(t--) {scanf ("%d",N); Init (n+1); Coun=1; for(intI=0; ii) {GetChar (); scanf ("%c%d"

Analysis: Use priority queues.Take next as priority, small first-out teamSort after reading the data, initialize the first element of the queue (0,a[0],0)Each time out of the team an element, the queue (sum,sum+a[nextid+1],nextid+1), (next,next+a[nextid+1],nextid+1), that is, whether add a[nextid+1] are considered in.In this way each new element is the next smallest (next), and the M-time is small.#include

[Transfer]: http://blog.csdn.net/linulysses/article/details/5771084 　　 Monotonous queue Assume that the sequence {XI} N= X1, X2,..., XnThere is a ordinal relation defined in } NIs {XI} NXJ1, Xj2,..., XjkWhere, J1 . This property is the monotonicity of the monotonic queue: the monotonicity between the subject and the elements. Similar to a general queue, the eleme

The slag is also in the effort to brush the white, although the basic of each problem can not be done, but each question is able to understand, and then achieve their own, and then write a report, I hope this method can be useful. This problem gives the coordinates and weight of n garbage, then the maximum load of the robot is C, when picking up rubbish must be picked up in the order of garbage. This problem comes up I think of backpack, either pick up, or go back, this method is actually can, b

the Dijkstra of the adjacent table and the priority queue. The only difference between queue and priority_queue is that in the priority queue, elements are not arranged in the order of entering the queue, but in the order of priority. Pop () deletes the element with the highest priority, not necessarily the element th

construction process of the Huffman tree: Get the two least numbers in all numbers each time and add them together to get the value of a non-leaf node. The two least numbers obtained before next time cannot be obtained, then, the values of non-leaf nodes can be obtained. For example, 3, 4, 5, add 3, 4 to 7, and then take 5, 7 (3, 4 can no longer be obtained), and add 12, 7 + 12 = 19, that is, the request. This operation can be well implemented by the priority

ACM? ACM !, AcmThoughts on ACM From yesterday to now, WA has finally made the question AC several times. Suddenly, I felt very much and recorded my feelings at the moment. I always feel that if I have enough time, there will be no questions I cannot answer. If you are not able to answer questions for the moment, remember that when you learn the relevant knowledge

Gao, school organization to Shaoshan play, I did not go, this time while 51, finally went to my hoping at of Shaoshan. In fact, I know all the attractions are similar, but because the TV drama "Just classmate Junior", let me to Mao Zedong has a

Sliding Window Time Limit: 12000MS Memory Limit: 65536K Total Submissions: 36212 Accepted: 10723 Case Time Limit: 5000MS DescriptionAn array of sizeN≤106 is given to you.

Topic Background: Previous revision code: #include #include using namespace std; string getcnt (int); String Sumstring (String, string); char Inttochar (int); int Chartoint (char); int main () {int m; while (cin>>m) {coutlen2)? Len2:len1;

JavaScript queue, priority queue and cyclic queue, and javascript queue A queue is an ordered set that complies with the FIFO principle.Add new elements to the end of the queue and remove elements from the top.

We are still on our way to dream-Summary of the sixth ACM competition in Shandong Province, acm in Shandong Province The results of the competition reached expectations (Gold Medal), which is far from my expectation in terms of the process (the competition was too messy and I played poorly). Fortunately, my teammates resisted the pressure at the crucial moment, the consequences are really unimaginable. Warm