asteroids html5

DescriptionBessie wants to navigate she spaceship through a dangerous asteroid field inthe shape of an n x N grid (1 Fortunately, Bessie have a powerful weapon that can vaporize all the Asteroidsin any given row or column of the grid with a Single shot. This weapon was quiteexpensive, so she wishes to use it sparingly. Given the location of any theasteroids in the field, find the minimum number of shots Bessie needs to fireto eliminate all of the asteroids.Input* Line 1:two integers N and K, sep

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 d

Document directory Problem description Input Output Sample Input Sample output Asteroids! Problem description You're in space. You want to get home. There are asteroids. You don't have want to hit them. Input Input 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 separa

G-asteroids!Time limit:MS Memory Limit:32768KB 64bit IO Format:%i64d %i64u SubmitStatusDescriptionYou ' re in space.You want to get home.There is asteroids.You don ' t want-to-hit them.InputInput to this problem would consist of a (Non-empty) series of up to data sets. Each data set is formatted according to the following description, and there would be no blank lines separating data s Ets.A Single Data set have 5 components:Start Line-a single line,

Topic Portal1 /*2 Test Instructions: Each time a row or column of obstacles can be eliminated, the minimum number of times required. 3 Hungarian algorithm: The row and column as two sets, when there are obstacles connected with an edge, the problem is converted to the minimum number of coverage = = Binary Graph maximum match4 Introduction to Fun:http://blog.csdn.net/dark_scope/article/details/88805475 */6#include 7#include 8#include 9#include Ten using namespacestd; One A Const intMAXN = 5e2 +T

Link: Nyoj 237: Click Here ~~ Poj 3041 asteroids: Click Here ~~ Question: The two questions are the same, but the translation is different. A legend-level game expert played a small game in his spare time. In the game, there were many enemies in an N * n Square area, players can use bombs to blow up all enemies in a row or column. He is a kind of player who wants to play well in any game, so he decided to use as few bombs as possible to blow up

Title Link: http://poj.org/problem?id=3041Test instructions: There are planets on the n*n matrix that can only be placed in one row or column at a time to eliminate all the planets in the bank or column, and to eliminate the minimum number of bullets consumed by all planets.Solution: A binary graph that behaves as a vertex set, as another vertex set. The topic turns into the least selected point (x or Y), so that it is adjacent to all the edges from these points, which is actually the minimum po

Description Bessie wants to drive her ship through a dangerous minor group. the Group is an nxn grid (1 ≤ n ≤ 500). There are K planets in the grid (1 ≤ k ≤ 10000 ). fortunately, Bessie has a very powerful weapon that can eliminate all the minor planets in one row or column at a time. This weapon is very expensive, so she wants to use as little as possible. the positions of all the minor planets are given, and the minimum number of shooting attempts by Bessie can be taken to eliminate all the mi

HDU-1240-Asteroids! Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1240 3D BFS, similar to 2D #include

Pku3041-Asteroids View rows and columns as points Using the Hungary Algorithm # Include

++; - } - - 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