numbered checkerboard

The following methods are used to delete even rows: : G/^/+1 D It used: Gbobal command, gbobal command format as follows: : [Range]global/{pattern}/{command} The global command is actually divided into two steps: First scans all rows in the range specified by [range], marks the rows that match {pattern}, and then, in turn, executes the {command} command on the tagged row, if the tagged row is deleted in the command action for the previously matched row, Move or merge, the tag d

Topic: N team competitions, numbered for 0,1,2 .... N-1, known for their strength-contrast relationships, stored in a two-dimensional array w[n][n], W[i][j] 's value represents a stronger branch of the team numbered i,j. So W[i][j]=i or J, now given their order of appearances and stored in the array order[n], such as order[n] = {4,3,5,8,1 ...}, then the first round is 4 to 3, 5 to 8. .......Winners are pro

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. The input and output stream in the javase---a read stream that corresponds to multiple output streams. And the resulting fragmented files are numbered sequentially.

appearnumber of cycles higher than array lengthPhenomenon, what is this for? Because the problem is in the reverse loop, if the reverse loop to I, can not find the odd, then there will be no previndex assignment, the outer loop will continue to go on, and then there will be more cycles than the length of the array, fix the code, increase the judgment stop flag bit:static void Setleftsingleandrightdoubleone (int[] arr) {int num = 0; var previndex = arr. Length-1; for (var i = 0; I OK, the alg

Import Java.io.file;import Java.io.fileinputstream;import Java.io.filenotfoundexception;import Java.io.fileoutputstream;import Java.io.ioexception;import Java.util.properties;public class FileSpilte {/** * @param args * @throws ioexception *//* * File Cutter, * One read stream, corresponding to multiple output streams, and the resulting fragmented file has an ordered number */public static void Main (string[] args) throws IOException {File Destdir = new File ("Teampfile\\filepart"); File Fi

Drop-down menu how to connect to a database display numbered content and pictures, etc.

-Ticketnum: -Thread Name:thread -Ticketnum: +Thread Name:thread +Ticketnum: -Thread Name:thread -Ticketnum: +Thread Name:thread +Ticketnum: AThread Name:thread ATicketnum: atThread Name:thread atTicketnum: -Thread Name:thread -Ticketnum: -Thread Name:thread -Ticketnum: -Thread Name:thread -Ticketnum: -Thread Name:thread -Ticketnum: -Thread Name:thread -Ticketnum: inThread Name:thread inTicketnum: -Thread Name:thread -Ticketnum: toThread Name:thread toTicketnum: +Thread Name:thread +Ticketnum: -

Label:Select Row_number () over (the partition by Group field order by sort field) as serial number, which shows the sequence number that is not unique within the group,SELECT * FROM (select Row_number () over (Order by sort field desc) as rowid from table) as table where rowID between (@PageInde X-1) * @PageSize +1 and @PageIndex * @PageSize paged query, can be used for small and medium data sorting,The SQL grouping is numbered sequentially based on

In 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.-------------------------------------------------------------------#include#include#include#include#includeint MN;Char Str[105][105];int dir[8][2]={{1,1},{-1,1},{-1,-1},{1,-1}};in

number of pieces to be placed. N When 1-1 indicates the end of the input.The following n-line description describes the shape of the chessboard: N characters per line, where # indicates the board area. Represents an empty area (the data guarantees that no extra blank lines or blank columns appear).OutputFor each set of data, give a line of output, the output is placed in the number of programs C (Data Assurance cSample Input2 #.. #4 4...#. #.. #.. #...-1-1Sample Output21stThe code is as follows

ElseAns = (ans + num[n-cnt])%MoD; - return; to } +DFS (pos+1, CNT); - if(!visd[node[pos]. D] !Visc[node[pos]. C]) the { *Visd[node[pos]. D] = Visc[node[pos]. C] =1; $DFS (pos+1, cnt+1);Panax NotoginsengVisd[node[pos]. D] = Visc[node[pos]. C] =0; - } the } + intMain () A { the //freopen ("In.txt", "R", stdin); + init (); - while(~SCANF ("%d%d", n, M)) $ { $ node.clear (); -memset (Mark,0,sizeof(Mark)); -Node.push_back (Node (-1, -1)); the for(

top and left of the minimum value. But in the beginning did not take into account the status of the situation is covered, only opened an array WA for a long time. Open two arrays one in the initial state after a storage and transfer state can be. rib[i][j]=min (rib[i-1][j],rib[i][j-1]) +board[i][j]. AC Code:　　1#include 2#include 3 using namespacestd;4 Const intINF =2147483647 ;5 intN;6 intboard[ the][ the],rib[ the][ the] ;7 intMain () {8scanf"%d",N);9 for(intI=1; ii)Ten for(intj

(Maxval, Dp[x1][y1][x2][y2]); the } About } the } theDP[I][J][K][L] = max (Dp[i][j][k][l], Maxval + arr[i][j] +arr[k][l]); the } + } - } the //two path to reach Point [M][n], then the point of the two path must choose: [M][n-1],[m-1][n]Bayi intans = max (Dp[m][n-1][m-1][n], Dp[m-1][n][m][n-1]); the //for (int i = 1; I the //{ - //For (int j = 1; j - // { the //for (int k = 1; k

] +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)