nxt creations

Links: http://www.spoj.com/problems/LCS/Test instructions two strings of LCSThere's really nothing to say, the first time. Let's turn over the online tutorial to see it again.Another found that Baidu internal users to communicate the use of pictures in the Baidu snapshot.#include #include#include#includeusing namespacestd;#defineMAXN 2510000#defineMaxt maxn*2structsam_node{intPnt,len; intnxt[ -]; voidPrint (intid=-1) {printf ("-----------------\ n"); printf ("Sam id:%d\n", id); printf ("Next: \

fourth? Can we place the second a in this position? We can see that the strings from the last character to the third a are aba, and the strings from the first character to the second a are also aba. Are they not the same? At this point, you should probably understand how KMP jumps. We can record an nxt array. nxt [I] indicates the length of the longest prefix Suffix from the first character to the first ch

(int K) { for (int i=head;i;i=a[i].point" if ((--k) ==1// find precursor {a[i].point =a[a[i].point].point; // break ; }}Traversing a linked list: for (int i=a[head].point;i;i=a[i].point) cout2, doubly linked listAs the name implies, a doubly linked list is a linked list with two directions. Unlike a one-way list, each node in a doubly linked list stores not only a pointer to the next node, but a pointer to the previous node. Its advantage is that access,

, but no fruitFirst, by the y-coordinate, we assume that this line is infinitely low, when the answer corresponds to the maximum number of points in the middle of the next two same colorAnd then this line up, one-time Delete rowMaintain the position of the previous and the latter of the same color with a doubly linked listDelete, delete this point from the tree array and the linked list, delete each one, and count the adjacent and adjacent answers before the point.And then you have to assign Y a

] = Ednum ++ ;} /* The function returns-1, indicating that there is a negative edge weight loop, which indicates that it is impossible.If the function returns-2, it indicates that the distance from the start point to the end point is uncertain. Here, we use dis [N] = inf to judge, because the start value is INF.Otherwise, DIS [N] is returned, where the maximum distance from the start point to the end point is stored. */ /* Module 3 */ Int Spfa (){ Int I, top, cur, TMP,

line:the number of OC Currences of the word W in the text T.Sample Input3bapcbapcazaazazazaverdiaverdxivyerdian1#include 2#include 3#include 4#include 5 using namespacestd;6 intn,m,nxt[10005],kk,t;7 Charb[10005],a[1000005];8 ///This topic adds multiple matches on the base KMP. 9 ///It means that after we've matched the string, we're going to jump to the right place and continue looking forTen ///continue to use KMP's ideas. Some locations have alread

#include 2#include 3#include 4#include 5 #defineMAXN 255000*26 using namespacestd;7 8 structNode {ints[ -],pre; } SUF[MAXN];9 CharS[MAXN],T[MAXN];Ten intI,j,n,m,k,len1,len2,len,last =1, root =1, cnt =1, ans; One intSTEP[MAXN]; A - voidAddintNXT) - { the intnow = ++cnt, p =Last ; -Step[now] = Step[p] +1; - for(; P! =0 SUF[P].S[NXT] = =0; p = suf[p].pre) SUF[P].S[NXT] =Now ; - if(p = =0) Suf[now

#include 5 #defineMAXN 105000*26 using namespacestd;7 8 structNode {ints[ -],pre; } SUF[MAXN];9 CharS[MAXN],T[MAXN];Ten intI,j,n,m,k,len1,len2,len,last =1, root =1, cnt =1; One intSTEP[MAXN],NANS[MAXN],ANS[MAXN],F[MAXN]; A - voidAddintNXT) - { the intnow = ++cnt, p =Last ; -Step[now] = Step[p] +1; - for(; P! =0 SUF[P].S[NXT] = =0; p = suf[p].pre) SUF[P].S[NXT] =Now ; - if(p = =0) Suf[now].pre

=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}voidOut (inta) {if(a0) {Putchar ('-'); a=-A;} if(a>=Ten) Out (A/Ten); Putchar (A%Ten+'0');}Const intn=1000005;//Code begin ...inttrie[205][ -], top;CharStr[n], s[ the];BOOLVis[n];voidInit () {top=1; MEM (trie[0],0);}voidInsChar*s) { intRT, NXT; for(rt=0; *s; RT=NXT, + +s) {NXT=trie[rt][*s-'a']; if(!

/* struct ListNode {int val; struct ListNode *next; ListNode (int x): Val (x), Next (NULL) {}}; * *: Class Solution {public:listnode* deleteduplication (listnode* phead) {ListNode *dummy=new ListNode (-1); Set an extra head node ListNode *cur=phead, *pre=dummy; dummy->next=phead; while (Curcur->next) {if (cur->val==cur->next->val)//If there is a duplicate node { int val=cur->val; while (Curcur->val==val) cur=cur->next;

Test instructions: Given an adjacency matrix to get a graph, to determine if there is Hamiltonian circuit, if present, output path, otherwise output-1;Idea: Each point as head traverse once to find Hamiltonian path, see if there is Hamiltonian circuit; a point need to be a special sentence;#include #include#includeusing namespacestd;intn,m;intmm[1005][1005];intnxt[500010];intcnt[500010],ans;voidSolveintStartintnum) { intj,k; memset (NXT,-1,sizeof(

)) The complexity of the solution. A good question. Hope to be able to persist in the future of each game to complete the title. AC Code/************************************************************************* > File Name:pf.cpp > Author:z nl1087 > Mail: [email protected] > Created time: 46/11 16:36:14 2015 ************************************** **********************************/#include #include #include #include #include #include #include #include #include #include #define LL Long Lo

=0x3f3f3f3f; the Const DoublePI = ACOs (-1.0); * ConstLL INF =0x3f3f3f3f3f3f3f3fll; $ Panax Notoginseng intT, N, Q, Le, root; - LL x; the inta[ *]; + A structnode{ the intCNT; + intnxt[3]; - $ voidinit () { $CNT =0; -nxt[0] = nxt[1] = -1; - } the}t[maxn* -]; - Wuyi voidInsert (LL N) { the intnow =Root; - for(inti =0; I +; i++) { WuA[i] = n%2; -N/=2; About } $ for(inti =

Bsgs#include #include#include#include#include#definell Long Longusing namespacestd;Const intMaxint= ((1 -)-1)*2+1;inta,b,c;structhashmap{Static Const intHa=999917, maxe=46340; intE,lnk[ha],son[maxe+5],nxt[maxe+5],w[maxe+5]; intTop,stk[maxe+5]; voidClear () {e=0; while(top) lnk[stk[top--]]=0;} voidADD (intXintY) {son[++e]=y;nxt[e]=lnk[x];w[e]=maxint;lnk[x]=E;} BOOLCountinty) {intx=y%Ha; for(intj=lnk[x];j;j=

princess does not pick;Inquiry [2, 2]: color 2 flower Only one flower, the princess does not pick;Inquiry [2, 3]: Because the color 2 flower has two flower, the princess picked the color 2 flower;Inquiry [3, 5]: Color 1, 2, 3 flowers each, the princess does not pick.HINT"Data Range"For 100% of data, 1≤n≤10^6,c≤n,m≤10^6. Analysis:It looks like the team can do it, but the data range is for the card, so we consider another way of doing it offline.First for each flower, find the next occurren

B2OJ_1565_[NOI2009] Plants vs. Zombies _ topological sort + Maximum weight closureTest instructions: n*m plants, each plant has fractions (which can be negative), and can protect the location of the plant. Can only eat from the right to the left, and can not eat is being protected, you may not eat, to obtain the maximum score.Analysis: Connect each plant to a plant that can protect it. The source point is connected with a positive point, and a negative weight point is connected to a meeting poin

) - return true; - Else - return false; - } - BOOLjudge () in { - for(intI=0; ii) { to for(intj=0; jj) { + if(grid[i][j]=='#'!Vis[i][j]) - return false; the } * } $ return true;Panax Notoginseng } - voidInit () the { + grass.clear (); Amemset (Vis,false,sizeofvis); the } + intBFS (Node n1,node n2) - { $Queue Q; $memset (Vis,false,sizeofvis); - while(!q.empty ()) Q.pop (); - Q.push (N1); the Q.push (n2); - intdepthest=0;Wu

Nothing dry write a template.#include #include#include#include#include#include#includeusing namespacestd;intRead () {intx=0, f=1;CharC=GetChar (); while(c'0'|| C>'9') {if(c=='-') f=-1; c=GetChar ();} while(c>='0'c'9') x= (x1) + (x3) + (c^ -), c=GetChar (); returnx*F;}#defineN 510#defineM 1600#defineINF 1000000000intn,m,p[n],v[n],u[n],tot=0, t=-1;BOOLFlag[n];intD[n],cur[n],q[n];structdata{intTo,nxt,cap,flow;} Edge[m1];mapint,BOOL>F;namespacetree{intp

SCU-4503Board question, according to test instructions description is divided into three steps1. For each location-centered palindrome string, and the maximum palindrome string2. For each palindrome string the length of the XOR prefix and3. Ask for two different or prefix and XOR maximum (Xor-trie)Then use the quick power to calculate the JD, and then the size of the FJDTake a look at the trie. Insert a 0 first#pragma COMMENT (linker, "/stack:102400000,102400000")#include #include #include #incl

SCU-4494 Given two sequences, ask what is the difference or maximum of one value from each of the two sequences Think of the numbers in a sequence as a binary 01 string, and then into the trie tree.Then the number of the B sequence is also treated as a 01-string, and then in the trie tree from high to low greedy to findIf the current bit of the 01 string found in the B sequence is 0, find the son of the trie tree that is 1Or you can only go to the right son of 0, and vice versa, th