asteroids arcade

++; - } - - BOOLDfsintu) { - for(inti = Head[u]; ~i; i =Edge[i].next) { in intv =edge[i].to; - if(!Vis[v]) { toVIS[V] =true; + if(Match[v] = =-1||DFS (Match[v])) { -MATCH[V] =u; the return true; * } $ }Panax Notoginseng } - return false; the } + A intHungry (intN) { the intres =0; + for(inti =1; I i) { -memset (Vis,false,sizeof(Vis)); $ if(Dfs (i)) $res++; - } - returnRes; the } - Wuyi intMain

- /*--------------------------------------------------------------------------------------*/ - using namespacestd; - in Const intMAXN = -+Ten; - intun,vn; to intg[2*maxn][2*MAXN],LINKER[MAXN]; + BOOLUSED[MAXN]; - the BOOLDfsintu) * { $ for(intv=1; v)Panax Notoginseng { - if(g[u][v]!)Used[v]) the { +USED[V] =true; A if(Linker[v] = =-1||DFS (Linker[v])) the { +LINKER[V] =u; - return true; $ } $ } - } -

This topic, three-dimensional space on the BFS, give you the starting point and end point, see if you can find a way, O said to go, x means you can not go! ~Understand the problem, you can use the queue to implement BFS to solve.The following is the code for the AC:#include Hangzhou Electric acm1240--asteroids!~~ Simple BFS

=s.x+dir[i][0]; $ inty=s.y+dir[i][1];Panax Notoginseng intz=s.z+dir[i][2]; - if(x>=0x0y0z'X') the { +ss.x=x; Ass.y=y; thess.z=Z; +ss.t=s.t+1; -map[z][y][x]='X'; $ if(x==exy==eyz==ez) $ { -flag=1; -step=ss.t; the return ; - }Wuyi Q.push (ss); the } - } Wu } - } About $ - intMain () - { - Charstr[ the],ch[ the]; A + while(SCANF ("%s%d",

Learning network flow in ing ... As a beginner practice is essential ~ ~ ~ Composition method because the book is very detailed, so simply say The beam as the vertex of the graph, the asteroid as the edge of the connection vertex, the map, because the minimum vertex coverage is equal to the maximum matching of two graphs , so the maximum matching of the binary graph can be obtained.Adjacency Matrix, Dfs seek augmented path, Hungarian algorithmAdjacency Matrix: Complexity O (n^3)If using ad

Test instructions: Three-bit space with N-layer for shortest circuitAnalysis: It is obvious that the wide search, but also wa many times:1. Think only 3 layers2. No special case is discussed: starting point and end point3. The judgment end condition is placed! X inside4. The coordinates of the input are changed into X, Y, z after the z,x,y in order to read the question carefully and found that the y,x,z changed the A.In summary: Still did not seriously read the question understanding test instru

Asteroids!Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission (s): 2599 Accepted Submission (s): 1745 Problem DescriptionYou're in space.You want to get home.There are asteroids.You don't have want to hit them. InputInput to this problem will consist of a (non-empty) series of up to 100 data sets. each data set will be formatted according to the following description, and there will be no blank lines sepa

PortalTest instructions: To a n*n matrix, some lattice have obstacles, ask us to eliminate these obstacles, asked each time to eliminate a row or a column of obstacles, at least a few times. Analytical:Take each row and each column as a point on either side of the two-part graph.If a lattice has a barrier, it corresponds to the row and column edge.Select the fewest points so that all edges are overwritten.Minimum point overlay.--code1#include 2#include 3 #defineM (x, a) memset (a, X, sizeof (a))

];Chara[6];intN;intK1,k2,k3,e1,e2,e3;intto[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};structnode{intx, Y, Z intstep;};intGointXintYintz) { if(000'O') return 1; return 0;}intBFS () {node st,ed; QueueQ; St.x=K1; St.y=K2; St.z=K3; St.step=0; memset (Visit,0,sizeof(visit)); VISIT[K1][K2][K3]=1; Q.push (ST); while(!Q.empty ()) {St=Q.front (); Q.pop (); if(st.x==e1st.y==e2st.z==E3) {cout" "Endl; return 0; } for(intI=0;i6; i++) {ed.x=st.x+to[i][0]; Ed.y=st.y+to[

First, the speed is subtracted, and a is moved and B does not move, and if the velocity is 0, it will never intersect.Enumerates each point of a and each segment of B, calculating the time at which the three points collinear.The time is sorted, three points are made for each interval, and the intersection area is calculated using half-plane intersection.Pay attention to the case where the intersection area is 0 but there are intersections.Time complexity $o (N^4\LOG^2N) $.#include 　　BZOJ4107: [W

; Q.push (to); } } }return 0;}intDfsintSintMax_vale) {intret=0, TMP;if(s==n*2+1)returnMax_vale; for(inti=head[s];i!=-1; i=edge[i].next) {intto=edge[i].to;if(dep[to]!=dep[s]+1|| edge[i].val==0)Continue; Tmp=dfs (To,min (max_vale-ret,edge[i].val)); edge[i].val-=tmp; edge[i^1].val+=tmp; ret+=tmp;if(Ret==max_vale)returnMax_vale; }returnRET;}intDinic () {intret=0; while(BFS (s,t)) { while(intT=dfs (S,inf)) {ret+=t; } }returnRET;}intMain () {init ();scanf("%d%d",

Test instructions: In a n*n matrix, there are m points, each elimination of a row or a column of points, the minimum number of operations;Idea: The x-axis and the y-axis as two different sets, each row or column as a point, each given point as an edge, so that the two-part map is built;Eliminate all points, the matrix is completely covered, so as to the minimum point coverage problem;Minimum cover number = maximum number of matches, so the maximum number of matches is required;#include #include#

The main idea: in an n * n grid, there are m obstacles, and now you have a weapon, this weapon can eliminate any line or a row of obstacles, now requires all obstacles to eliminate, ask at least how many times the use of this weaponProblem-solving ideas: Think of long time how to solve the problems of row and column ... How to represent two point setsFinally suddenly thought, since do not know how to deal with the ranks, the ranks are divided into two points, the point on behalf of the relations

Not happy, simple three-dimensional BFS#include HDU 1240 Asteroids!

Question address: Asteroids Question: Give you a grid of N * n. In the coordinates of the given stars, the operation is to eliminate a whole row or a column of stars each time and ask you the minimum number of operations required. Solution: The key is to think about binary matching. You can draw a picture first, and create the coordinate rows and columns of the star position, (1-> 1, 1-> 3)/(2-> 2)/(3-> 2 ). how can we minimize the elimination of all

Click Open Link Konig theorem:Minimum vertex overwrite in a bipartite graph = maximum number of matching in a Bipartite Graph N * n networks contain K minor planets. The position of the minor I is (Ri, CI ). Today, there is a powerful weapon that can use a single beam to destroy an entire row or a whole column of minor planets. How many light beams are required to use this weapon to destroy all the minor planets? Analysis: a bipartite graph is created based on the coordinates of the left and ri

Tags: Io OS for SP C on amp R size The maximum matching of a bipartite graph = the minimum vertex overwrite (Konig theorem) = the maximum matching of the largest independent set the maximum matching of the classic three models. This is the minimum vertex overwrite, by the way, my experience with this question is that you must be careful and then be familiar with all the details of the template during debugging !! # Include # Include # Include Using namespace STD; Const int maxn = 10005;

vertex overwrite = maximum number of matches. Therefore, the question is convertedMaximum matching of a Bipartite Graph, It is a perfect solution. By the way, the zju1002 bipartite graph model cannot directly use rows or columns as the vertices of the Bipartite Graph, but needsSplit point. For more information, see my other solution report. Http://blog.csdn.net/tiaotiaoyly/archive/2008/10/23/3122893.aspx /* Pku3041 Asteroids */ # Include #

Common 3D wide search, please note thatInput:Columns, rows, and layers # Include # include # include # include # include # define M 11 using namespace STD; int dir [6] [3] = {0, 0}, {0,-}, {, 0}, {-, 0}, {0, 0, 0, 1 },{,-1 }}; // six

4.2.3 System Management Functional Group The System Management function group is the core part of the backend server software, which is composed of several modules, such as "field Bureau Generation Control", "Game road singles print", "Historical