uva timecard

The numbers on the bricks can eventually be seen as the linear combination of the last line, with a maximum of 9 independent elements.In the general practice of this type of problem, the linear combination can list the equation and then the Gaussian elimination element.For this problem, just determine the last line of the remaining 4 variables, for the last line of J position, it on the position of a number and the number of contributionsThe number of schemes equal to the path to that location c

The main idea: in a non-descending sequence of length n, there are M queries, each query (I,J) indicates the number of occurrences of the most frequently occurring element in the interval (i,j).Title analysis: Because the sequence is ordered, you can segment the sequence and record the number of elements per paragraph, the segment num (i) each element belongs to, the left endpoint L (i) and the right endpoint R (i) for each element. So for each query:Ans (i,j) =max (max (R (i)-i+1,j-l (j) +1), R

Topic PortalTest instructions: Two dogs running on a polyline, the speed is unknown, at the same time to go, at the same time. The difference between the maximum distance and the minimum distance between two dogs during a runAnalysis: Training Guide P261, consider the relative motion, set a stationary, B relative a motion, the relative motion vector: vb-va (can be understood as the speed vector), then is the PA to the line segment Pb-pb+vb-va distance maximum/************************************

, finally found that the order he gave is that the first event is in the number of occurrences, as in the previous sort of problem,,, alas:-(, to kneel, can be such a son, And then changed the code, submit, AC, (⊙o⊙) Oh, it's not easy ah, a longest common sub-sequence of living written for two hours ~~~~~~~~~~~~~ slowly are tears ah ~~~~~~~~~~~~~~!Test instructions probably says, give you 1 to n events, sorted by the sequence of the first row and the longest common subsequence after sorting by t

Scriber,Then to the second scriber etc. But each scriber must is assigned at least one book.1#include 2#include 3 #defineM (a) memset (A,0,sizeof (a))4 Long Longa[510],b[510];5 intMain ()6 {7 Long LongI,j,k,m,n,p,q,x,y,z,t,l,r;8scanf"%lld",t);9 while(t--)Ten { One M (a); A M (b); -scanf"%lld%lld",n,p); - for(i=1; i) thescanf"%lld",a[i]); -x=0; -y=a[1]; - for(i=1; i) + { -x+=A[i]; + if(a[i]>y) y=A[i]; A } atL=y; -R=x; - while(lR

Test instructionsThe relationship of the tree is given in the form of a stringAnalysis:Direct DFS Search, line I, J if it is a letter, Judge I+1 Line J is ' | ' If you find the first '-' in line i+2, find the letter in line I+3 and repeat.Code:#include #include #include #include using namespace Std;const int maxn=210;int n;Char A[MAXN][MAXN];void Dfs (int r,int c){printf ("%c (", a[r][c]);if (r+1{int i=c;while (i-1>=0a[r+2][i-1]== '-')i--;while (a[r+2][i]== '-' a[r+3][i]!= '){if (!isspace (A[r+3

UVA 3*n+1 problem.Solution: M,n (MThe maximum number of steps is obtained by calculating the M to n sequentially.Note: 1. Enter the format.Use while (scanf ("%d%d", m,n)!=eof){} 2. The output is the same as the problem. 3. The selection of the variable type when the boundary is 1The code is as follows:#include #includevoidFindresult (intMin,intMax) { inti,count,temp; Temp=0;//maximum number of steps for temp storage Long LongN//Note that the N

number of columns, please find a layout such the length of the longest contiguous spaces Between words is minimum. Figure I.6:a Good and the most beautiful layout. InputThe input consists of multiple datasets, each in the following format.W NX1x2 ... XNW, N, and XI is all integers.W is the number of columns (3≤ W ≤80,000).n is the number of words (2≤ N ≤50,000). XI is the number of characters in the I-th word (1≤ xi ≤ (W? 1)/2). Note that the upper bound in XI assures that th

The direct LCS is the time Complexity O (p*q), but the sequence elements are different, as long as one of the sequences is mapped into an orderly,Another sequence that does the same mapping, without the direct deletion, becomes the one that asks for another sequence of Lis.#include using namespacestd;intRead () {CharC while(C=getchar (),c'0'|| C>'9'); intRe = c'0'; while(C=getchar (), c>='0'c'9') Re = re*Ten+c-'0'; returnre;}Const intMAXN = -* -;intMP[MAXN];intS[MAXN],G[MAXN];//#define LOCALint

The main idea: to give a sequence of 1~n, each operation can be the length of an even-numbered sequence to exchange the first half and the last half of the position. Find the steps to turn this sequence into ascending order.Problem Analysis: Structural solution.The code is as follows:# include　　UVA-1611 Crane (construction)

integer so is the size of the largest clique in T (G).Sample input15 51 22 33 14 15 2Output for sample input4 Zachary FriggstadMain topic: T Group test Data. Give a picture of the graph G. Find the node set with the largest node number, so that at random two nodes in the node U and v meet: either u can reach v. Either v can reach U (U and v mutually reachable also can). Problem Solving Ideas: "The points in the same strongly connected component are either selec

The main topic: with n length may not be equal to match the maximum number of sticks.Title Analysis: Enumerate the lengths of all possible equal-length sticks, match each root by DFS, and prune them in the process. The length of the stick is sorted from large to small, that is to say, it is always preferred to match each long stick. The pruning scheme is as follows: 1. If the i-1 stick is not used in the matching of the current scheme and length[i]==length[i-1], then the root of the first stick

to perform m operations on each of these boxes for one of the x, Y reverse order of the two to the left of Y.After the operation is complete, all odd digits of the original box sequence number and;Direct simulation will definitely time out with the list in the STL also time-out can only use the array itself to simulate a doubly linked list le[i],ri[i] respectively, the box to the left of the first box and the right box in the ordinal code has a stare#include Copyright notice: This article Bo M

)); ECNT=0; while(m--){ intU,v,w; scanf"%d%d%d",u,v,W); Addedge (--u,--v,w); Addedge (V,U,W); } Dijkstra (--s,--e,d1); Dijkstra (E,S,D2); intPick =-1, p2, ans =D1[e]; intK scanf"%d",j); for(inti =0; i ){ intU,v,w; scanf"%d%d%d",u,v,W); if(D1[--u] + W v]) {ans= D1[u] + W +D2[v]; Pick= u; P2 =v; }Else if(D2[u] + W D1[v]) {ans= D2[u] + W +D1[v]; Pick= V; P2 =u; } } intu; if(~pick) {Stackint>Stk; U=pick; Stk.push (U); while(U! =S) {

(inti =1; I 0; for(intU =0; U ){ intV0 = Sccno[u]; wei[v0]++; for(inti = Head[u]; ~i; i =Nxt[i]) { intV1 =Sccno[to[i]]; if(V0! = v1) G[v0].push_back (v1), deg[v1]++; } }}intDP[MAXN];inttopo () {memset (DP,-1,sizeof(DP)); Queueint>Q; for(inti =1; I ){ if(!deg[i]) Q.push (i), dp[i] =Wei[i]; } while(Q.size ()) {intU =Q.front (); Q.pop (); for(inti =0; I int) G[u].size (); i++){ intv =G[u][i]; DP[V]= Max (dp[v],dp[u]+Wei[v]); if(--deg[v] = =0) Q.push (v

Test instructions: Similar to the 8 queen question, except that each lattice on the board has a score and the Queen's position is the one that can be obtained. Or the rule of the 8 Queen's question, and finally get the best possible score.Idea: 8 Queen problem solution, after 8 Queens to judge the score on the line.In the If statement judgment content that piece unexpectedly still write wrong once, not satisfied ~You can also use vis[3][2*8]; array markers, and then quicklyCode:#include

Test instructions: n planets, given a one-dimensional coordinate of each point, can delete m points, so that all remaining points to the center of the remaining points of the sum of the square and the smallest, to find the minimum value;Idea: Ruler method, each maintenance n-m point; Long Long is read in with%LLD, but codeblocks compile will cause problems, so use cin>>;#include #include#include#include#includeusing namespacestd;intt,n,m;Long Longs1,s2;Long Longnum[50500];Doublesum,mid,cnt;intMa

Title: Give a string of O and x (length 1-80), statistical score.The score for each o is the current number of consecutive o, for example, the ooxxoxxooo score is 1+2+0+0+1+0+0+1+2+3Problem-solving ideas: Using a variable term records the current O scores, if there is O, then term+1, if X appears, then term=0;Then use a sum to record the sum, no time add term can/*UVa 1585 Score---water problem*/#include#includeConst intMAXN = -;intMain () {intT; Char

Title: Two people take turns in the n*n of the parallelogram lattice into black and white two-color pieces,If the black side can give the creation of a continuous line from 1~n lines then black wins, otherwise white wins.Analysis: Graph theory, search. Use DFS or FloodFill solution to find solutions that can reach the low end from the top.Description: Target 600 question ╮(╯▽╰)╭.#include UVa 260-il gioco dell ' X

Manacher (); - getd (); About for(intI=0; i) $ { - for(intj=0; J) -Dp[i]=min (Dp[i], (d[i][j]==0?1:d p[d[i][j]-1]+1)); - d[i].clear (); A } +printf"%d\n", dp[len-1]); the } -}Then see LRJ code, Memory search, some violence. Before I heard that the VIS does not have to be initialized every time, this is the first time to see how to write.int Kase; { .... if return p[i][j]; = Kase; ....} int Main () { ... .. 0 sizeof (Vis)); for