witch puzzle

Title Link: http://poj.org/problem?id=1651Idea: In addition to the top and bottom two numbers can not be taken, all the number is required to take out, each take a number to spend this number and the two adjacent number of products, the cost of the cost is minimalUse Dp[i][j] to denote the minimum cost of the interval [i,j], then there is Dp[i][j]=min (Dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]) i+11#include 2#include 3#include 4#include 5#include string.h>6#include 7#include 8#include 9#include Ten#inclu

Test instructions: slightly.Analysis: First of all assume that the former n-2 has been put, then put Nth, first consider what n-1 put, then there are two situations.If the n-1 is the same as the 1th, then the nth can be placed on the basis of N-2 2, that is 2 * f (n-2), that is, because N-1 and 1th, as the same,So there are two types of nth (not the same as the 1th kind). So what if the first n-1 and the 1th one? Then the nth is the same as the first n-1, there is no choice but to put one.The co

? means the scaanf and cin of the flapping street. ② then it entered the instruction is infinite and then to 0 end, well this is still relatively easy but actually the command in the middle there will be empty line ah I go ... There may be more than one line of instructions, what does a blank line mean? It means that getline also jumped on the street. ③ Fortunately we also have C=getchar () while loop, but the output data also has the first regular pit is the letter finally can not have the extr

);#defineOut Freopen ("OUT.txt", "w", stdout);using namespacestd;Const intinf=0x3f3f3f3f;Const intmod=1e9+7;Const intn=5e4+Ten; LL N,s;vectorint>Eg[n];intLinker[n];BOOLVis[n];BOOLDfsintu) { for(intI=0; I) { intto=Eg[u][i]; if(!Vis[to]) {Vis[to]=1; if(linker[to]==-1||DFS (Linker[to])) {Linker[to]=u; return 1; } } } return 0;}voidAddintu) { inttmp=s/u+1; for(LL i=tmp*u; iS].PB (u);}intMain () {#ifndef Online_judge//inch#endif intt,cas=0; scanf ("%d",t); while(t--

Test instructionsGiven the 0/1 matrix of n * m, then the Q group asks for the maximum side length of the full 1 squares contained in the asked rectangle.ExercisesFirst consider the brute force algorithm, for each inquiry of the rectangle, in which the violence found all 1 squares, now the question is how to find the full 1 square, can be recursive implementation, DP[I][J] to (i, j) as the bottom right corner of the full square longest side length.Dp[i][j] = min (Dp[i-1][j], dp[i][j-1], dp[i-1][j

integer A that satisfies this condition.1#include 2#include 3#include 4 using namespacestd;5 6 intMain ()7 {8 intK,a;9 while(cin>>k)Ten { One if(k% $==0) A { -printf"no\n"); - Continue; the } - for(a=0;a $;++a) - { - if((a*k)% $== -) + { -printf"%d\n", a); + Break; A } at } - if(a== $) printf ("no\n"); - } - - return 0; -}View Code(H

(t) c[++len]=T; - } + A BOOLPD () at { - if(l>p)return 1; - for(intI=1; i) - if(Ans[i]return 1; - return 0; - } in - voidChuinty) to { + inti=len,t,x=Z[len]; - for(l=0; i;) the { * if(i>1) {if(XElse Break; $d[++l]=x/y;x=x%y;Panax Notoginseng } - if(x>=y) d[++l]=x/y; the if(PD ()) for(p=l,i=1; iD[i]; + } A the intMain () + { -scanf"%d%d%d", n,a[0],b[0]); $ for(i=1; i"%d%d", a[i],b[i]), s[i]=a[i]*b[i],id[i]=i; $Sort (id+1, id+1+N,CMP);

; - } the - for(intI=1; I//find the maximum valueWuyiAns =Max (F[i], ans); the } - Wuprintf ("%d", ans); - About return 0 ; $}PS: This problem is visualized as the extension of the longest descending subsequence in two dimensions. By the way, the code for the memory search is attached as follows:1#include"bits/stdc++.h"2 3 using namespacestd;4 Const intMAXN = the ;5 6 intH[MAXN][MAXN], F[MAXN][MAXN];7 intN, m, ans =1;8 9 intDFS (intXint

overlap with the left nodeAnd it is possible that the left lever length of the right node is greater than the left side's lever length, and vice versa.Therefore, this special case needs to be considered when generating the parent node to calculate the total length of the left and right balances.#include #include#include#includeusing namespacestd;Const intMAXN = A;intN, IDX, F[MAXN], W[MAXN];DoubleR, ans, RW[MAXN], LW[MAXN];voidBuild_binary_tree (intDEP) { if(dep = = N)return; for(inti =0; i

http://blog.csdn.net/libin56842/article/details/9747021Http://www.cnblogs.com/devil-91/archive/2012/06/26/2562976.html#include #includestring>#include#include#include#include#include#include#include#include#include#include#includeSet>#include#includeusing namespacestd;#defineMem (A, B) memset (A,b,sizeof (a))#definePF printf#defineSF scanf#defineSPF sprintf#definePB Push_back#defineDebug printf ("!\n")#defineINF 10000#defineMAX (A, b) a>b?a:b#defineBlank pf ("\ n")#defineLL Long Long#defineAll (

With Gobang, the three-child chess is in the nine lattice can achieve three pieces of chess together to winDid not modify the wording, the implementation of the function but the cumbersome and a lot of unsound place #include2. Modified after the three sub-chess programThis article is from the "incomparable Warm yang" blog, please be sure to keep this source http://10797127.blog.51cto.com/10787127/1710502Three sub-chess game of puzzle fun

/*It's not a difficult question, but it's a matter of the dictionary tree, but space is a problem. Start writing such a struct node{int next[27],sum[27]; bool over;} T[MAXN]; It's not going to work. The ER opened up the MLE asked Wmy very witty said with map works and then the card space to see them using pointers to dynamically allocate memory but I'm not very good at it .*/#include#include#include#include#defineMAXN 480010using namespacestd;inttopt;Chars[ -];structnode{Mapint,int>next,sum; BOO

Test instructions: Give a sequence, select a number at a time (recorded as B), calculate the nearest number to his left (recorded as a), the right distance from his nearest number (recorded as C), calculate the A*b*c,sum+=a*b*c repeat operation until the end of the first two numberDifferent schemes correspond to different sums to calculate the minimum sum valueAnalysis: Typical interval dp,dp[i][j] represents the minimum value of the selected number of all from I to J.Dp[i][j]=min (dp[i][j],dp[i

Consider a multiple of a number in turn, a multiple of two digits (LCM), a multiple of three numbers (LCM) ...It will be found that there is a rule, odd number of time to add the number of cases, even the number of numbers to subtract the number of cases.A type of only 10 numbers, which can be enumerated in binary.#include #include#includeusing namespacestd;intn,m,save[ -];Long Longans;intMain () { while(~SCANF ("%d%d",n,M)) { for(intI=0; i) {scanf ("%d",Save[i]); } ans=0; for(intI=1;i1)

two cases, either just well found, one on the leftLike what2 3 4 5 7 find 3 return 2Find 6 return 5 We don't know if it's on the left or the right. The difference is small, can be considered inFind 8 to return to the last position#include //#include #include string.h>#include#include#include#includeSet>#include#include#include#include#include#defineMAXN (100000+10)#defineMAXM (100000)#defineINF 0X3FFFFFFF#defineINF 0x3f3f3f3f#defineLson l,m,rt#defineRson m+1,r,rt#defineLL Long Long#defineULL un

) *2p0;//selected two other sons by two times P0.。。。 (There are n-1 items altogether)Add up is (1+...+n-1) * (Son_num[p]-son_num[x])/(n-1)/n;i.e. (Son_num[p]-son_num[x])/2.0;And because it will bring a root node, so to reduce one more;i.e. (son_num[p]-son_num[x]-1)/2.0;Perfect ~ Then on the code.#include #include#include#include#include#includestring>#include#include#includeSet>#includeusing namespacestd;inta[100100];vectorint>edge[100100];intnum_son[100100];Doubleans[100100];intDfs_fir (introot

327typedefBOOLBoolean;8Ifstream Fin ("scifest.in");9Ofstream Fout ("Scifest.out");Ten Boolean found; One intM,n; A intbuf; -typedefstructmydata{ - intPEO; theUnsignedintactivities[Ten]; - }mydata; -MyData *s; -Unsignedintac[Ten]; +Boolean CMP (Constmydata A,Constmydatab) { - returnA.peo >B.peo; + } A Boolean init () { at if(! (fin>>n>>m))return false; -s =Newmydata[(Const int) (n +1)]; -Found =false; - for(inti =0; I ){ -S[I].PEO =0; -memset (S[i].activities,0,sizeof(s[i].activ

; + stringstr; -mapstring,int>table; + intdir[]= {-1,1,-4,4}; Move Direction A voidBFS () at { -QueueQ; - stringS,s_next; - P temp; - intCur,next; -Q.push (P ("01234567",0)); intable["01234567"]=0; - while(!q.empty ()) to { +temp=Q.front (); - Q.pop (); thes=Temp.first; *Cur=Temp.second; $ for(intI=0; i4; i++)//move 0 in four directionsPanax Notoginseng { -next=cur+Dir[i]; theStr=s; + swap (Str[cur],str[next]); A if(next>=0next8! (cur==3next==4) ! (c

//J also indicates the number of times a player can attack before the tower hits J soldiers. * for(inti =1; I ){ $ for(intj =0; J +; j + +){Panax NotoginsengF[i][j + Xbs[i].kill +1] = max (F[i-1][J], f[i][j + Xbs[i].kill +1]); - if(J >= Xbs[i].pkill-Xbs[i].kill) { theF[i][j-(xbs[i].pkill-xbs[i].kill)] = max (F[i-1][J] + +Xbs[i].money, F[i][j-(Xbs[i].pkill-Xbs[i].kill)]); A } the } + } - intMAXV =

;}voidDanceintK) {if(K+f () >= ans)return;if(! r[0] {ans = min (ans, k);return; }intCol = r[0]; for(inti=r[0]; i!=0; I=r[i])if(S[i] for(intI=d[col]; I!=col; I=d[i]) {remove (i); for(intJ=r[i]; J!=i; J=R[J]) Remove (j); Dance (k +1); for(intJ=l[i]; J!=i; J=L[J]) Resume (j); Resume (i); }}intMain () {intN, M; while(~SCANF ("%d%d", n, m)) {intnum =0; for(intI=1; i for(intj=1; j"%d", g[i][j]);if(G[i][j]) p[j+ (I-1) *m] = ++num; }intR, C; scanf"%d%d", r, c); Init ((n-r+1) * (m-c+1),