nxt creations

;//If you do not specify GreaterintN//Number of buttonsintans[maxn]={0};//Record Storage Results//Initialize input graphvoidinit () {scanf ("%d",N); G=New int*[n+5]; for(inti =1; I //The edge of each node { intm =0; scanf ("%d", m);//The number of edges that record this pointG[i] =New int[m+Ten]; g[i][0] =m; for(intj =1; J 0]; ++j) {scanf ("%d",G[i][j]); inch[g[i][j]]++;//the entry of the J-Point plus one } }} //returns whether the topology can be sortedBOOLTopologicals

componentThe non-direction graphSide Double Unicom diagram: If in an undirected graph, any two points have at least two edges not repeating the path, it is said that the graph is the edge of the two-connected .Side Dual-Link component: The maximal sub-graph of the side-double connectivity is called the edge-connected component .The principle is similar to the method of strong Unicom component.1 voidTarjan (intUintFA) {2Dfn[u] = Low[u] = + +tn;3St[++top] =u;4Vis[u] =true;5 for(intI=head[u];

A * algorithm, also called Heuristic Search, is to design an estimate function, and then in the process of searching for an orderly search, we set to the current state of the cost of f (x), to the target state of the estimated cost of H (x), then we sort by H (x) +f (x), the beginning to the current distance f (x ), the current to the end of the shortest circuit is g (x), and then a violent search can be. --by Vane#include using namespacestd;Const intn= -;Const intm=10050;Const intinf=1e9;intn,m

limit of the X composition.Code#include using namespaceStd;typedefLong Longll;Const intm= -+3;Const intn= -;Const intinf=0x3f3f3f3f;intn,m;intL[n],p[n],c[n];intNd[n];BOOLBa[n];structnode{intNxt,to;} e[2*N];inthd[n],cnt;voidAddintXinty) {e[++cnt].nxt=Hd[x]; E[cnt].to=y; HD[X]=CNT;}voiddpintx) { if(Ba[x]) {l[x]=min (l[x],m/c[x]); return; } for(intI=hd[x];i;i=e[i].nxt) { inty=e[i].to; DP

Blog:www.wjyyy.top ac automata is a convenient multi-pattern string matching algorithm for cancer . Based on the dictionary tree, a similar kmp thinking is used. ac automata unlike KMP, AC automata can match multiple pattern strings at the same time, and the complexity does not reach too high. If the string is matched multiple times with KMP, the complexity is \ (O (k (n+m)) \). We know that it is most convenient to use a dictionary tree to match a string header to match the other strings ex

Topic links\ (description\)Each point has a cost Si and value pi, which requires selecting some connected blocks with a root, the total size is k, making \ (\frac{∑pi}{∑si}\) the largest\ (solution\)01 Fractional plan, then DP, set F[I][J] indicates the maximum weight of the I subtree selected J and, direct violent backpack transfer can beWhen enumerating the number of child nodes selected, assume X has 1.2.3.4 four child nodes with a complexity of \ (1*sz[1]+sz[1]*sz[2]+ (sz[1]+sz[2)) *sz[3]+ (

", X) One #definePL (x) printf ("%lld\n", X) A #defineRep (I, A, n) for (int i=a;i - #definePer (i, a, n) for (int i=n-1;i>=a;i--) - #defineFi first the #defineSe Second - using namespacestd; -typedef pairint,int>PII; - Const intN = 1e5 +5; + Const intMoD = 1e9 +7; - Const intMOD =998244353; + ConstDB PI = ACOs (-1.0); A ConstDB EPS = 1e-Ten; at Const intINF =0x3f3f3f3f; - Constll INF =0x3fffffffffffffff; - - intn,m,cnt; - BOOLmp[205][205]; - intDu[n],head[n]; in inta[205]; - structp{intTO,

3198: [sdoi2013]springtime limit:40 Sec Memory limit:256 MBsubmit:1143 solved:366[Submit] [Status] [Discuss] DescriptionInputOutputSample Input3 31 2 3 4 5 61 2 3 0 0 00 0 0 4 5 6 Sample Output2 HINTDragonite Correcting Data SourceHash simple , but hash is a bit troublesome.Ans= at least the logarithm of K *c (K,K)-at least k+1 logarithmic *c (k+1,k) + at least k+2 logarithmic *c (k+2,k) ...For this C (i+k,k), my understanding of this logarithm was computed C (k+i,k) timesHash judg

[Topic link]http://poj.org/problem?id=3417AlgorithmDifferential on treeCode#include #include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#includeSet>#include#include#include#includestring>#include#include#include#include#include#includeusing namespacestd;#defineMAXN 100010#defineMaxlog 20structedge{intTO,NXT;} E[MA

maintain, so consider the insertion of the point of destruction,when you insert point now[i], it means that now[i + 1].....now[k] has been inserted (not yet destroyed), and Now[1]........now[i-1]has been inserted (destroyed), so you can use and check set, to maintain the unicom block. #include using namespacestd;Const intMAXN =400005;intN, M, Fa[maxn], K, VIS[MAXN], ANS[MAXN];intHEAD[MAXN], cnt =1, tot =0, NOW[MAXN];structnode{intV, NXT;} G[MAXN];voi

#include using namespacestd;Const intMAXN = 3e5 +Ten;structedge{intV, NXT;} EDGE[MAXN];CharCH[MAXN];intDEG[MAXN];intdp[maxn][ -];intHEAD[MAXN], CNT;intN, M;inlinevoidinit () { for(intI=0; i) {Head[i]= -1, deg[i] =0; for(intj=0; j -; J + +) Dp[i][j]=0; } CNT=0;} InlinevoidAddedge (intFrom,intTo ) {EDGE[CNT].V=to ; EDGE[CNT].NXT=Head[from]; Head[from]= cnt++;}#defineDebug 0intMainvoid){#ifDebugFreopen ("In.t

++;} InlineintRead () {CharC=NC ();intx=0, f=1; while(c'0'|| C>'9'){if(c=='-') f=-1; c=NC ();} while(c>='0'c'9') {x=x*Ten+c-'0'; c=NC ();} returnx*F;}intn,m,s,t;intH[MAXN];//the height of each nodeintF[MAXN];//traffic that can flow from each nodeintGAP[MAXN];//the number of each heightstructnode{intU,V,FLOW,NXT;} EDGE[MAXN];intHEAD[MAXN];intnum=0;//Note here num must start from 0InlinevoidAdd_edge (intXintYintz) {edge[num].u=x; EDGE[NUM].V=y; Edge[n

Bzoj_1179_[apio2009]atm_tarjan+spfaTest instructions: http://www.lydsy.com/JudgeOnline/problem.php?id=1179Analysis:Obviously there is no way to direct the longest road, then shrink the point to run.The bar connects to meeting point.Code:1#include 2#include string.h>3#include 4 using namespacestd;5 #defineN 5000506 inthead[n],to[n1],nxt[n1],val[n],can[n],cancan[n];7 intn,m,bel[n],dfn[n],low[n],st[n],top,scc,tot,cnt,s,t;8 intIns[n],sum[n],x[n],y[n],q[n]

integer) as stringIf W> = 0 then mimeencode = mid $ (base64, W + 1, 1) else mimeencode = ""End Function Private function mimedecode (A as string) as integerIf Len (A) = 0 then mimedecode =-1: Exit FunctionMimedecode = instr (base64, a)-1End Function Public Function encode (byval indium as string, byval e as long, byval N as long) as stringDim s as stringS = ""M = IndiumIf M = "" Then exit functionS = mult (clng (ASC (mid (M, 1, 1), E, n)For I = 2 to Len (m)S = S "+" mult (clng (ASC (mid (M, I

Practiced hand#include #include#include#include#defineMAXV 100500#defineMaxe 200500using namespacestd;Long LongN,q,x,y,z,g[maxv],nume=1, dis[maxv],top[maxv],fath[maxv],son[maxv],size[maxv],dfn[maxv],mx[maxv],times=0;structedge{Long LongV,NXT;} E[maxe];Long Longls[maxv2],rs[maxv2],val[maxv2],lazy[maxv2],sum[maxv2],tot=0, Root;Chars[Ten];voidAddedge (Long LongULong Longv) {e[++nume].v=v; E[NUME].NXT=G[u]; G[u

#defineSe Second#defineFS First#defineLL Long Long#defineCLR (x) memset (x,0,sizeof x)#defineMC (x, y) memcpy (x,y,sizeof (×))#defineSZ (x) ((int) (x). Size ())#definefor (It,c) for (__typeof ((c). Begin ()) it= (c). Begin (); it!= (c). end (); it++)#defineLson l,m,rt#defineRson m+1,r,rttypedef pairint,int>P;Const Doubleeps=1e-9;Const intn=5e5+Ten;Const intm=1e3+Ten;ConstLL mod=1000000007;Const intinf=1e9+Ten;intN,cnt,num;inthead[n],dp[n][2],ans[n],to[n];structedge{intV,

Chain water problem.#include #include#include#include#defineMAXV 100500#defineMaxe 200500#defineINF 2147483647using namespacestd;intN,m,type,x,y,z,val[maxv],id=1, g[maxv],nume=0;intw[maxv],fw[maxv],dis[maxv],fath[maxv],top[maxv],size[maxv],son[maxv],mx[maxv],times=0;introot,tot=0,ls[maxv2],rs[maxv2],mn[maxv2],lazy[maxv2];structedge{intV,NXT;} E[maxe];voidAddedge (intUintv) {e[++nume].v=v; E[NUME].NXT=G[u];