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Problem Description:In a checkerboard of 2^k*2^k squares, there is a grid that differs from other squares, called special squares.Now invite 4 different L-shaped dominoes to cover all squares of a given chessboard except for special parties.For example: This is a k=2 when the board, the special point coordinates (0,1).Four L-type dominoesProgram code:#include Executes the result."Recursive" checkerboard Ove

Title Description DescriptionThere is a 5x5 chessboard with some of the squares dyed black and the other squares white, and your task is to dye the board with a few squares so that all black squares can be connected together and you have the fewest number of squares to stain. Read into the state of an initial checkerboard, and the output needs to be dyed at least as many squares as possible so that all black squares are connected together. (Note: Con

Topic Link http://poj.org/problem?id=1191problem: 1191User: yinjianMemory: 568K time: 16MSLanguage: C + + Result: AcceptedProblem Solving Report:1, the formula can be used to simplify the mathematical method, is to find the number (sum) of the sum of the squares and the smallest.2, each division has four kinds of cases (recursion).3, each time the location of the division to be compared, so as to find the best.#include #include#include#includestring.h>using namespacestd;ints[9][9];//the score o

C-Checkerboard IssuesTime limit:MS Memory Limit:10000KB 64bit IO Format:%i64d %i64u SubmitStatusPracticePOJ 1321DescriptionIn 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 i

Checkerboard Issues Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26726 Accepted: 13242 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 th

Title Link: Checkerboard IssuesParsing: DFS violence searches from top to bottom, from left to right.AC Code:Code 1#include Code 2#include Copyright NOTICE: This article is sxk original article, reprint please attach this article link ^_^ POJ 1321 Checkerboard Problem (DFS)

Title Link: POJ 1321 checkerboard Problem Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 28825 Accepted: 14276 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 an

Checkerboard Issues Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24958 Accepted: 12333 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 th

2017-08-26 20:18:50Writer:pprpThe question probably describes:Have a 2k? 2k checkers, just one square is black, the other is white. Your task is to cover all white squares with an L-shaped card containing 3 squares.Black squares cannot be overwritten, and any white squares cannot be overwritten by two or more cards at the same time.Divide and conquer the solution, when the division of the time to determine the state, need what state, the end of conditionsThe end condition is that when siz

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