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(2-Point Matching of graph theory algorithms) hdu 1281 (board game) A: give an n * m board and put it on the board. The cars that are placed on the Board cannot attack each other (they can attack each other on the same line or column), and only some dots can hold the

rows=15;//checkerboard row number public static final int cols=15;//Checkerboard Number point[] chesslist=new point[(rows+1) * (cols+1)];//initial each array element is null Boolean isblack=true;//the default start is black first Bo Olean gameover=false;//whether the game ends int chesscount;//The number of current

Board GameTime limit:2000MS Memory Limit:65536KB 64bit IO Format:%i64d %i64 U Gym 100935GDescriptionStandard Input/outputStatementsFeras bought to he nephew Saleem a new game to help him learning. The game consists of a board with 4 rows and 4 columns with cubes. Every cube has a

Title Description Description The game on a board with a 10*10 lattice, the initial pieces in the upper left corner, the end point is the lower right corner, the chessboard of each lattice has a number 0 to 9, each time a pawn can move to the right or lower adjacent lattice, to find a number and the smallest and throug

Board gamesTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 2758 Accepted Submission (s): 1613Problem description and Gardon play a game: to a n*m chess board, in the lattice to put as much as possible some chess inside the "car", and so they can not attack each other, which is of course very simple, but Gardon

(inti =1; I if(Book[i] = =0 P[u][i] = =1) {Book[i] =1;if(Match[i] = =0|| DFS (Match[i])) {Match[i] = u;return true; } } }return false;}intMain () {intCA =1; while(scanf("%d%d%d", n, m, k)! = EOF) {memset(P,0,sizeof(p));memset(Match,0,sizeof(match)); for(inti =0; I scanf("%d%d", a, b); it[i].x = A; It[i].y = b; P[A][B] =1; }intAns =0; for(inti =1; I memset(Book,0,sizeof(book));if(Dfs (i)) ans++; }intTMP =0; for(intj =0; J 0;intANS1 =0;memset(Match,0,sizeof(match));

Problem Solving Ideas:Enumerates all the squares on the board, and if the point is deleted, the maximum number of matches is reduced, then the point is the key point.#include HDU 1281 board game (two-part chart matching)

Problem description James and gardon are playing a game: For an N * m board, place as many "cars" as possible in the grid as possible in chess ", this makes it easy for them not to attack each other, But gardon restricts that only some grids can be placed. xiaoxi easily solves this problem (see) note that the locations where vehicles cannot be placed do not affect the mutual attack of vehicles. So gardon wa

http://acm.hdu.edu.cn/showproblem.php?pid=1281Board gamesTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 3096 Accepted Submission (s): 1827Problem description and Gardon play a game: to a n*m chess board, in the lattice to put as much as possible some chess inside the "car", and so they can not attack each other, which is of course very simple, but Gardon

Analysis: The problem can be used to match the x-coordinate of the y-coordinate, the match is a successful one can put a piece of points, the last to find the best binary matching is the maximum number of pieces can be put. The maximum match of the binary graph is using the Hungarian algorithm. After the deletion of an edge to determine whether a point is a key point, if the deletion, the maximum number of

Online board game 6930 that can be played in linux ~~ -- Linux general technology-Linux technology and application information. For details, refer to the following section. I don't want to talk about it anymore. Http://www.6930.com In addition: the webmaster Income Plan is coming soon. 5 RMB will be sent if the number of recommenders playing in over 100 gam

- if(map[u][v]!Used[v]) the { -used[v]=true; - if(linker[v]==-1||DFS (Linker[v])) -{//find an augmented path, reverse +linker[v]=u; - return true; + } A } at return false;//don't forget it, always forget it. - } - intHungary () - { - intres=0; - intu; inmemset (linker,-1,sizeof(linker)); - for(u=1; u) to { +memset (Used,0,sizeof(used)); - if(Dfs (U)) res++; the } * returnRes; $ }Panax Notogi

Analysis: First to find out the maximum number of matches, and then with a matching point of a single removal to see if the maximum match, can be the key point (very violent AH), but unexpectedly just 31ms************************************************* #include #include string.h>#include usingnamespaceStdConstintMAXN = the;BOOLG[MAXN][MAXN], USED[MAXN];BOOLFind (intIintNintP[]){ for(intj=1;j{if(G[i][j] used[j] = =false){USED[J] =true;if(!p[j] | | F

The labels of the rows and columns make up the x and Y sets, and then if the x, Y coordinates can be placed in the car, a line is connected between the Xs and y points, and no peers and trains do not attack each other, so finding the maximum match is the maximum number of cars that can be placed and then enumerating each point to determine whether it is an important#include HDU 1281 Board

HDU 1281 board game (two-point matching and edge deletion) Based on the topic description, what is important? After finding the maximum match, perform enumeration and delete the edge in sequence to see if the maximum match number will change. If it changes, this is an important point. # Include # Include # Include # Include #