nxt creations

of investment 6, profit is 10, so the maximum gain of 4. "Scoring method" is not part of the subject, your program output only and our answer exactly the same to obtain full marks, otherwise do not score. Data size and convention 80%: N≤200,m≤1 000. 100% of the data: N≤5 000,m≤50 000,0≤ci≤100,0≤pi≤100.SOURCE Analysis:Bare Maximum right-closing sub-graph ...Each transit station as a negative power point, customer groups as a positive point, and then each customer to the transfer station to conne

mid ... TAT ...Code:1#include 2#include 3#include 4#include 5 //by Neighthorn6 #defineINF 0x3f3f3f3f7 #defineINF 0x3f3f3f3f3f3f3f3f8 using namespacestd;9 Ten Const intmaxn= $+5, maxm=100000+5; One A intn,m,s,t,cnt,sum,a[maxn],b[maxn],hd[maxn*2],fl[maxm],to[maxm],nxt[maxm],pos[maxn*2]; - Long LongDis[maxn][maxn],max; - theInlinevoidAddintSintXinty) { -fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++; -fl[cn

Topic Link: hdu_5963_ FriendTest instructionsChinese, not explainedExercisesTake a look at the sample, and you'll see if X is the node that wins, or whether the Benquan value associated with x is the same or odd.1#include 2 #defineF (I,A,B) for (int i=a;i3 using namespacestd;4typedefLong Longll;5 6 Const intn=80007;7 intt,n,m,g[n],nxt[n],v[n],w[n],ed;8 9 voidAdgintXintYintZ) {v[++ed]=y,w[ed]=z,nxt[ed]=g[x],

update the time to find all the direct son of U set to v OK, can be updated directly with F[V][CU], but if the nearest center of V is not CU, but is CV, then in the subtree of V to enumerate all CVs to find a minimum of F[V][CV], with f[v][cv]+k to update, Then why do you add K now? Since the CV is in the subtree of V, the center of U is not a CV but a CU, and it is a certainty that you have this CV built as a center, and you must add K.You can then use the Floyd to process the distance between

the list to store the position of all the puddings for each color, merge the linked list and color markers each time it is modified, record the link list length, and the list of violent modified lengths.However the writing hung up. After bowing to the puzzle, it was found that the original color of the pudding was not recorded, causing the stain to deviate.↓ Note the statement differences between lines 37 and 38. To determine whether to dye according to the primary color.1#include 2#include 3#i

() or listen (). This segment is discarded, but RST is not returned. The client tries to re-establish the connection request. Create a new socket: When a socket in the listening status receives this segment, a sub-socket will be created, and socket {}, tcpcb {} And pub {} will also be created. If an error occurs, the corresponding socket is removed and the memory is released through the flag. The TCP connection fails. If the cache queue is full, TCP considers an error to occur and all subseque

The tree array is more elegant than the line tree... We have done a good job in anti-prime number tabulation, that is, the formula for pushing the next position from the current position has not been developed, so we can see other people's... If (P [now]. x> 0) NXT = (NXT-1 + P [now]. X) % leave; Else NXT = (NXT + P [n

],vis2[max];intpnt[max*2],head[max],nxt[max*2],e;//if the number of sides is wrong, the report time-out what ghost ... InlinevoidAddedge (intUintv) {Pnt[e]=v;nxt[e]=head[u];head[u]=e++;}intMaxx[max],minx[max],limit;BOOLDfsintu) { intv; Maxx[u]=inf;minx[u]=0;//Each virtual node range is set to an infinite size if(Vis1[u]) maxx[u]=minx[u]=A[u]; for(inti=head[u];i!=-1; i=

Supposedly a very simple bfs.I was just beginning to write a two-way BFS, and the results never went away.And then I thought, if the search speed is different, there will be a bug.I had to change it into two separate BFS + last enumeration to determine the minimum value1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intd[4][2]={1,0,-1,0,0,1,0,-1};7 structnode{8 intt,x,y;9 }s,e;Ten intvis1[205][205],vis2[205][205]; One Charmap[205][205]; A intn,m; -Queue QF,QB; - BOOLChec

5Maximum flow of network streamsIf the maximum flow is equal to M, then there is a solution that iterates through all the sides of the question, and records the answer.1 /*by Silvern*/2#include 3#include 4#include 5#include 6#include 7#include 8#include 9 using namespacestd;Ten Const intinf=1e9; One Const intmxn=1050; A intRead () { - intx=0, f=1;CharCh=GetChar (); - while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} the while(ch>='0' ch'9') {x=x*Ten+ch-'0'; ch=GetChar ();} -

...There's really nothing to write about ...However, it is important to think that it is possible to use the +∞ side to express the flow of traffic ...For each pigsty we connect a volume from S to each pigsty for the side of the pig quantity, from each customer to the t even a volume for the side of the desired adoption quantity, for each guest, if this guest is the first time to open the guest of this pigsty, then the pigsty to this guest even +∞ side, Otherwise, the last guest who opened the

Test instructions: There is an n-point m-side of the direction of the graph, each point has its own weight, can be negative, now from a point to go, until you can not go to other points, each through a point, you may choose to get or not to get its weight, each point can go multiple times, but the right value can only be obtained once, Ask how many weights you can get at the end.Each point can go more than once, the weight can only be obtained once, passing the time of the weight may not be obta

children the exact time she'll spend on the roads. Can you help her?InputThe first line contains three integers n, q, S. There was n huts in XX Village, q messages to process, and wind was currently in Hut S. N The following n-1 lines each contains three integers a, B and W. That means there was a road directly connecting hut A and B, time required is W. 1The following q lines is one of the following and the types:Message a:0 u A kid in Hut U calls wind. She should go to hut u from hers current

*headb) { if(Heada = = NULL | | headb = =NULL)returnNULL; intAddressa;//fin of A. intADDRESSB;//fin of B. intLenA = Countlength (heada,Addressa); intLenB = Countlength (headb,ADDRESSB); if(Addressa! = ADDRESSB)//if has a intersect returnNULL; ListNode*tmpheada = (ListNode *) Addressa;//To store Heada ' s tail for reverse.Reverselink (Heada); intNewlenb = Countlength (headb,ADDRESSB); intTonew =Findcount (LENA,LENB,NEWLENB); Reverselink (Tmpheada); returnFi

the KMP algorithm, it is a matrixSo that we can count them without a heavy leak.Another matrix acceleration: F[i][j] The method is a linear homogeneous recursion, can be constructed into a matrix, and then add a speed.This problem is going to leave a hole ... Because I don't know qaq.1#include 2#include 3#include 4#include 5#include 6#include 7 #definePAU Putchar (")8 #defineENT Putchar (' \ n ')9 using namespacestd;Ten Const intmaxn= -; One intN,MOD,LEN,NX