nxt shooterbot

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

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=

) - 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

Enumeration left interval segment tree maintain maximum value#include #include #include #include #include #include #include #include #include #include #define LL Long Long#define MOD 999911659 //2 3 4679 35617 #define N 5000051using namespace STD;intSC () {intI=0, f=1;CharC=getchar (); while(c>' 9 '|| c' 0 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 'c' 9 ') i=i*Ten+c-' 0 ', C=getchar ();returnI*f;}Long LongMx[n],tag[n],ans;intp[n/5],nxt[n/5],n,m,

integer Wi indicates the weight of the i-th.Then M lines follow. Each line represents a query. It contains integers L and R, meaning the king wanted to ask about the situation if he sent all rabbits from the L-th one to the r-th one into prison.(1 The input ends with n = 0 and M = 0.OutputFor every query, output one line indicating the answer.Sample Input3 22 1 41 21 36 43 6 1 2 5 31 34 64 42 60 0Sample Output211312HintIn the second case, the answer of the 4-th query are 2, because only 1 and 5

Let ' s Orz Yts//#pragma comment (linker, "/stack:20240000,20240000")#include #include #include #include #include #include #include #include #include #include #include #define N 500005using namespace STD;BOOLmark[n][ A];inthead[n],nxt[n1],lst[n1];intf[n][ A],g[n][ A],w[n];intN,d,m,tot;voidInsertintXintY) {lst[++tot]=y; nxt[tot]=head[x]; head[x]=tot; Lst[++tot]=x; Nxt