memory matrix game

Description Q is a very smart child. In addition to chess, he also enjoys playing a computer puzzle game, matrix game. Matrix games are played in an N * n black/white matrix (just like playing chess, the color is random ). You can perform two operations on the

DescriptionLittle Q is a very clever kid, besides chess, he also likes to play a computer puzzle game-matrix game. The matrix game is carried out in a n*n black-and-white phalanx (as in chess, except that the color is random). You can do two operations on the

2016-05-31 17:26:45Title Link: NOIP 2007 matrix Fetch game (CODEVS)Main topic:Given a matrix, each time at the beginning of each line or at the end of the line takes a number multiplied by the number of 2^, to get the most points obtainedSolution:Dynamic planningDP[I][J] Indicates the maximum score obtained from the current line I position to the J positionTransf

"Title description"Handsome often play with classmates a matrix take number game: for a given n*m matrix, each element of the matrix A[i,j] areis a non-negative integer. The rules of the game are as follows:1. Each fetch must take one element from each row, a total of n. Aft

(C.val[i][j] >= MoD) C.VAL[I][J]-=MoD; } } returnC; }}; Mat Pow_mod (Mat A, LL b) {mat ret (A.SZ); for(intI=1; i1; while(b) {if(B 1) ret = RET *A; A= A *A; b>>=1; }returnret;} LL Cal (intNintMintk) {MAT A (M+1); for(intI=1; i1][i] = 1LL * k * (K-1); for(intI=2; i1] = k *1LL; A=Pow_mod (A, N); LL ret=0; for(intI=1; i) ret= (ret + a.val[i][1]) %MoD; returnret;}intMainvoid) {__sttime (); __ioput (); intncase; SCI (ncase); while(ncase--){ intN, M, K; SCIII

DescriptionDescription [Problem description]Shuai often plays a matrix fetch game with his classmates: for a given n * M matrix, each element of the matrix AIJ isIt is a non-negative integer. The rules are as follows:1. Each time the number is obtained, one element must be taken from each row, N in total. Aft

"BZOJ1444" [Jsoi2009] Interesting game descriptionInputAttention, 0OutputSample InputSample OutputHINT30% of data guaranteed, n≤2. 50% of data guaranteed, n≤5. 100% data guarantee, N, L, m≤10. Solution: The practice of the subject is really a lot ah, probability DP, expect DP, of course, and the moment by black technology ~is to run the AC automaton first, make the transfer matrix, and then squared 50 times

Test instructions: n * N 01 Matrix, can do two operations: ① two lines exchange; ② two columns. Ask if you can go through several operations, so that the matrix from the top left to the bottom right diagonal is all 1. The problem: The first line Exchange, the column exchange when the relative position of the element will not change, the matrix is equal to the exi

; i)Panax Notoginseng { -Ret.uni= ((ret.uni*Ten)% (mod-1) +str[i]-'0')% (mod-1); theRet.ord= ((ret.ord*Ten)%mod+str[i]-'0')%MOD; + } A returnret; the } + - voidInit () $ { $ CharSTRM[MAXN],STRN[MAXN]; -scanf"%s%s", STRN,STRM); -n=get_value (STRN); them=get_value (STRM); -scanf"%lld%lld%lld%lld",a,b,c,d);Wuyi } the - voidGet_ans () Wu { -ll F;//=f[n+1][1] About if(a==1) $ { -ll D= ((((C* (m.ord-1))%mod) *b)%mod+d)%MOD; - if(c==1) f= (1+N.ORD*D)%MOD; -

http://www.lydsy.com/JudgeOnline/problem.php?id=1059 (Topic link)Test instructionsA 01 matrix that can exchange two or more columns arbitrarily, asking if the main diagonal is all 1 after several exchanges.SolutionHzwer: Peers with the same column no matter how many times the transformation still peers or the same column, so the title can be converted to find n different rows or the same column of points.Codebzoj1059#include　　"bzoj1059" zjoi2007-

a, and a [n-1] is not met, because the n times cannot be passed to a. The other type is that the n-1 times are not in the hands of, the nth time the ball was held and then passed to Jia had a [n] method, according to the addition principle a [n-1] + A [n] = (K-1) ^ (n-1) A [1] = 0 is used because a is the serving operator. Idea: An (N indicates the number of times that the ball is passed n times and the number of times that the ball is returned to the armor plate ); A1 = 0; A2 = (K-1) ^ 1-a1; A

];i;i=next[i])if(bo[zhi[i]]!=t) A { thebo[zhi[i]]=T; + if(!pre[zhi[i]]| |find (pre[zhi[i]],t)) - { $pre[zhi[i]]=x; $ return 1; - } - } the return 0; - }Wuyi intMain () the { - Inin (t); Wu while(t--) - { AboutInin (n); Clear (PRE,0), Clear (head,0); ed=0; Clear (Bo,0); $Re (I,1, N) Re (J,1, N) - { - Inin (A[i][j]); - if(a[i][j]==1) Add (i,j); A } + intans=0; theRe (I,1, N)if(Find (i,i)) ans++; -

[Topic link]https://www.lydsy.com/JudgeOnline/problem.php?id=1059AlgorithmMaximum matching of binary graphsComplexity of Time: O (n^3)Code#include using namespacestd;#defineMAXN 210structedge{intto, NXT;} E[MAXN*MAXN];intn, tot;intMATCH[MAXN],HEAD[MAXN];BOOLvisited[maxn];templatevoidChkmax (T x,t y) {x =Max (x, y);} TemplatevoidChkmin (T x,t y) {x =min (x, y);} TemplatevoidRead (T x) {T F=1; x =0; Charc =GetChar (); for(;!isdigit (c); c = GetChar ())if(c = ='-') F =-F; for(; IsDigit (c); c = G

Reprint Please specify source: http://blog.csdn.net/vmurder/article/details/42884845First we can use a monotone stack to sweep through each row, maintain a point to the top and bottom of each can extend how long.Of course, this can be done, is also very disgusting.We can sweep through each line to maintain the maximum left-to-right distance at the current [column height] of each point.Of course, a certain point can be lowered a little bit, it will be wider, the result is better, but obviously si

Notice how to change it, but if you put a black square on the diagonal, the black squares of its original row will not work.So to choose the N group of not heavy row and column combination, here with the Hungarian algorithm to do two-figure matching (with time stamp optimization)#include Bzoj 1059: [ZJOI2007] Matrix game "Hungarian algorithm"

Test instructions: Finding the largest sub-matrixAnalysis: Directly using the formula of the most large matrix to do time-out. A change of mind, the problem with the previous question 1506 has a relationship, first of each layer to calculate the maximum height of each element can reach, and then with 15,061 samples. The DP staging data is used for height and area two locations. For the height of the double cycle, and then each layer for the bottom are

Portal: http://www.lydsy.com/JudgeOnline/problem.php?id=1059Save the Hungarian template.#include 　　_bzoj1059 [ZJOI2007] Matrix game "Two-point graph matching"

Game: tianlong BabuVersion: 0.13.0402System: Windows XPTool: ce5.2 + od1.10Objective: To search for the character base address Step 1: search for the person Hp with Ce, get a bunch of addresses, continue searching after blood loss, get the unique address 0abdc360 (HP address) Step 2: Switch the map and find that the value in the address is no longer HP, it is a dynamic address. Repeat the first step to search for a new HP address (the address is omitt

Share the fate, comments, @author: White robe path, when the bitter harmPath code:1, because the path here Blog directory does not own the whole, for the time being with the essay directory structure, so the two-level directory that is ignored. Header format is roughly (original or turn) level two directory (title)2, because the view and the previous record too messy, so only a little manual removal (recall, sorting). Welcome to discuss, knowledge and ability is always asked out is not (hehe, so

, Vis, DIR, NX, NY) +1); Vis[nx][ny]=false; } } if(MAX = =-1)return 1; Dp[x][y]=MAX; returnDp[x][y]; } intLongestincreasingpath (vectorint>>matrix) { intn =matrix.size (); if(n = =0)return 0; intm = matrix[0].size (); Vectorint> > DP (N, vectorint> (M,0)); VectorBOOL> > Vis (N, vectorBOOL> (M,false)); Vectorint> > Dir (4); dir[0].push_back (-1); dir[0].push_back (0); dir[