checkerboard dimensions

] +arr[i][j]);Wuyi } the if(Check (x2, y2, I-1)) - { WuDP[I][J] = max (Dp[i][j], Dp[x2][y2] +arr[i][j]); - } About } $ } - /*for (int i = 1; I - { - For (int j = 1; J A { + cout the } - }*/ $cout 1) Endl; the } the the intMain () the { - #ifdef HOME inFreopen ("inch","R", stdin); the //freopen ("Out", "w", stdout); the #endif About for(inti =0; i i) the { the for(intj =0; J j) the

instead of some bfs thought of Dfs, but because there is no storage state like BFS, so do some repetitive work, time will be slower than the first method.) ）3) WA, not yet debugged:Because it is important to realize that the key to solving the problem is that the sub-change and change two times in each position is actually the same. change the effect of paragraph A and then change the sub-B is equivalent to first change the sub-B and then change the sub-A is not related to the order, so I envi

pieces we need to place, and then ask us how to put them in several ways. First we can make it clear that this is a topic of deep search, similar to the eight Queens question. We create a function Dfs is used to accumulate the number of feasible scenarios, we walk through a column we will mark it down the next time we can not be placed in this column (because the title is not allowed to be placed in the same row and the same column)Then start looking for a viable place from the next line, until

(Visi,0,sizeof(Visi)); memset (VISJ,0,sizeof(VISJ)); memset (Vis1,0,sizeof(VIS1)); memset (Vis2,0,sizeof(VIS2)); if(Dfs (0,0,0)) Break; } printf ("Case %d:%d\n",++Cas,maxd); } return 0;}The first kind of pruning3. You can place rows (or columns) on a row-by-line basis. There is also a pruning is up to 5, so maxd==4 has no solution, direct output 5.0.201s#include #includeConst intMAXN = One;CharG[MAXN][MAXN];intMaxd;intn,m;BOOLvisi[maxn],visj[maxn],vis1[maxn1],vis2[maxn1];BOOLDfsintDintsi) {

Test instructions: Given a n*n (nAnalysis: Use a one-dimensional array to mark rows and columns and search deep.#include #includeChars[Ten][Ten];intN, M;intvis[Ten];intans;voidDfsintCurintStep) { if(step==m) {ans++; return; } if(cur>n-1)return; for(intI=cur; i) for(intj=0; j) if(s[i][j]=='#'!Vis[j]) {Vis[j]=1; DFS (i+1, step+1); VIS[J]=0; }}intMain () { while(SCANF ("%d%d", n, m)) {if(n==-1m==-1) Break; for(intI=0; i) scanf ("%s", S[i]); memset (Vis,0,sizeof(VIS)); Ans=0; DFS (0,0

So simple a topic, thought for so long, also WA once, unforgivable ah, more and more feel stupid, Dfs search will not write ... LuoConfusion, we must seriously think about the topic, seriously write code:Code:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. POJ 1321 Checkerboard Problem Simple DFS

One, test instructions:Chinese questionsSecond, analysis:The main use of compressed DP and memory search ideasThree, code:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. POJ 1191 Checkerboard Segmentation (compression dp+ memory Search)

Simple Search Practice Backtracking1#include 2#include 3#include 4#include 5#include string>6#include 7#include 8#include 9#include Ten#include One#include A#include -#include - using namespacestd; the intn,k; - intans; - Chars[Ten][Ten]; - intusex[Ten],usey[Ten]; + BOOLOkintXinty) { - if(x0|| x>=n| | y0|| y>=N) + return 0; A if(s[x][y]=='.'){ at return 0; - } - if(usex[x]==1|| usey[y]==1) - return 0; - return 1; - } in voidDfsintXinty) { - if(

Reprint Please specify the source Thank you: http://blog.csdn.net/vmurder/article/details/42886393ExercisesWell, this picture is really not very good to do,But we can convert it to perfection is a 1/0 sub-matrix problem.is to reverse the point 01 of the same row parity, and then it's OK (this is obviously, need to prove the message).And then we're asking for the maximum sub-matrix.As for the square? To seek the sub-matrix by the way, that is the ans1,This is obviously because we enumerate the du

Topic: Given a black and white matrix, the largest black and white sub-squares and sub-matricesReverse the point of an odd position, then the largest sub-square and sub-matrix of the solid colorThe matrix is divided into layers, above each layer is a similar to the ▆▃▇▂▉, with a monotonous stack run out each point left to right can expand to the maximum distance, update the answer can be#include Bzoj 1057 ZJOI2007 checkerboard making monotone stack

Title: Give a matrix formed by 01, ask the matrix of the largest area of the square and the rectangle, any one of the squares adjacent to a different lattice.Idea: In fact, all (i + j) 1 in the position of the number XOR, it becomes all 0 or 1 of the largest squares and rectangles. The first question is water DP, the second question can be monotonic stack or hanging line. are very well written.CODE:#include Bzoj 1057 Zjoi 2007 checkerboard Production

Chinese problem, title no longer explained, simple DFS;#include #include #include #include #include BOOL Maps[10][10];//maps make a chessboard, judge this point can play chess maps[0][i] used to record column I can play chessint N, k, ans, Hang;//hang represents the moment in the first few lines of chessvoid DFS (int k){if (k = = 0) {//pieces are put outans++;Return}if (Hang > N)//The number of rows checked is greater than or equal to N, at which point K is not equal to 0;Returnif (N-hang + 1 Re

][j][k]+=dp[i-1][k][j-1]* (k) * (m-j-k+1)/2;5: Put a gun in a row with a gun and a gun in a row with a cannon, the equation is dp[i][k][j]+= (k+2) * (k+1)/2*dp[i-1][k+2][j-2];6: Put a gun in a row without a gun and put a gun in a row without a gun, the equation is dp[i][k][j]+= (m-j-k+2) * (m-j-k+1)/2*dp[i-1][k-2][j];Okay, now we can put the code on.　　#include #include#defineMoD 999983using namespacestd;intn,m;Long Long intdp[101][101][101],ans=0;intMain () {scanf ("%d%d",n,m); dp[0][0][0]=1; f

Title Description Given a 4*4 01 chessboard, 1 represents a pawn, 0 represents a space, and a pawn 1 can be moved to a space next to or below four positions in each of the adjacent places. Then given your target chessboard, ask you at least how many steps you can take to turn the current chessboard into a target checkerboard state. input The first line enters an integer t, which indicates that there is a T group of test data. Next, we give a 4*4 of o

topic AC.Idea: Store the Board area position in an array, and then each checkerboard area either places a piece, or does not put a pawn, deep search (equivalent to generate the number of permutations).For a chessboard area can play chess pieces, like the eight queens, we use two Boolean array to maintain, r[i] to indicate whether there is a pawn in line I, C[j] Indicates whether there is a pawn in column J.The deep search function maintains two param

DescriptionIn a given shape of the chessboard (the shape may be irregular) on the top of the pieces, chess pieces no difference. If you need to place any of the two pieces in the same row or column in the chessboard, please program the chessboard with a given shape and size, and put all the feasible arrangement C for the K pieces.InputThe input contains multiple sets of test data.The first row of each set of data is two positive integers, n K, separated by a space, indicating that the chessboard

[i]) { - for(intj =1; J //look for "#" in the line to if(Map[i][j] = ='#' Row[j] = =0) { +Chose[i] = j;//The first t piece is placed in the J position of line I -ROW[J] =1; theChessdfs (t +1); *ROW[J] =0; $ }Panax Notoginseng } - } the } + } A } the + intCheckintk) { - for(inti =1; I ) { $ if(Map[k][i] = ='#' Chose[k])return 1; $ } - return 0; - } the - voidLinedfs (intt) {Wuyi

if((I-1) >=0) Dp[i][j] + = dp[i-1][j]; - if(J-1) >=0) Dp[i][j] + = dp[i][j-1]; the } + } A } thecout Endl; + } - $ intMain () $ { - -CIN >> n >> m >> x >>y; thememset (arr,0,sizeof(arr)); -Memset (DP,0,sizeof(DP));Wuyi //Remove the other horse's control points the for(inti =0; i i) - { Wu inttx = x +Avoidx[i]; - intTy = y +Avoidy[i]; About if(TX >=0) (TX 0) (Ty m)) $ { -Arr[tx][ty] =true; -

answer is to see the board on the odd number of or even a number of black pieces;AC Code:#include /*#include */using namespacestd;#defineRiep (n) for (int i=1;i#defineRIOP (n) for (int i=0;i#defineRJEP (n) for (int j=1;j#defineRJOP (n) for (int j=0;j#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong LongLL;ConstLL mod=1e9+7;Const DoublePi=acos (-1.0);ConstLL inf=1e18;Const intn=1e5+4;intn,m,a[ the][ the];intMain () {intT; scanf ("%d",t); while(t--) {scanf ("%d%d",n,m); intsum=0; Riep (n)

DescriptionIn a given shape of the chessboard (the shape may be irregular) on the top of the pieces, chess pieces no difference. If you need to place any of the two pieces in the same row or column in the chessboard, please program the chessboard with a given shape and size, and put all the feasible arrangement C for the K pieces.InputThe input contains multiple sets of test data.The first row of each set of data is two positive integers, n K, separated by a space, indicating that the chessboard