csr2 racing

Topic Links:http://poj.org/problem?id=2287Main topic:Tian Bogey and King Horse racing, two people each have n horse, each horse has a speed, run fast wins, slow lose. Tian Bogey every win one200, flat, no money, lost. 200. Every time the king first, Tian bogey again. Q: What is the maximum amount of tin bogey? Ideas:Greedy thought. Now the field bogey and the King's horse to sort. Tian bogey horse speed from small to large arrangement, King's horse sp

Tian Ji -- the horse racing question: Tianji's horse racing problem is that the horse wins and loses based on the speed of the horse, and each game wins 200 points. input392 83 7195 92 74392 83 7092 91 60output0200 Strategy: 1. If Tian Ji's fastest horse is faster than Wang's, He will directly win the game. 2. If Tian Ji's fastest horse is slower than Wang's fastest horse, he will lose one

Description: The historical story "Tian Ji Horse Racing" in ancient China is well known to everyone. In other words, Qi Wang and Tian Ji have to race horses again. They each send n horses. In each game, the losing party will give the winning party 200 yuan or two gold coins. If it is a draw, both parties do not have to pay for it. Now the speed value of each horse is fixed and known, and Qi Wang does not care about the order of the horse. How can Tia

A racing competition was held in space. There are two flight modes: the first is to fly from a normal road, but only from a small number to a place with a large number; the second is to go directly, but it takes a fixed time. Ask how much time is required to traverse all vertices at exactly one time. Idea: billing flow. Split each vertex, s to the start point of each vertex, and the cost of connection to each endpoint is fixed, the original edge of t

Tian Bogey horse race, roughly test instructions is Tian Bogey and King Horse racing, win a innings 200 yuan, lose a game lose 200 yuan, draw property does not move.Enter an integer n, and the next line is the N horse of Tian Bogey, and the next line is the King's N horse. When n is 0 o'clock end. This is titled Greedy algorithm solution, there are two ideas. Thought two: 1, beginning is also the first sort, you can use sort quick row, 2, then the big

understand the following article.Let me start by stating that there are many ways to develop a car. I racing version after countless different methods, countless times after the final selection of a popular accepted. These methods, let me a way.First of all, you have to understand that the vehicle can not be a whole shell, because in contact with the ground, the force on the vehicle can not achieve your intended target, caused, must be in the vehicle

First, case analysis.Speed, racing source platform build rack q:2947702644We can think of, since it is slow, sometimes direct access can not, before is not a problem, and now suddenly there is a problem, it must be our nginx and PHP response to the cause, may be caused by the huge increase in the number of user connections to other domain sites. Then we can find out the root of the problem and optimize it. Then with their own experience and Baidu, to

Tags: save named out Port close EMC Zook. SH source1. Download Beijing-Racing platform Source Building2. Upload files to the Linux server via XFTP3. Unzip the zookeeper using the command tar--zxvf zookeeper-3.4.124. Configure the system environment variable Vim/etc/profileExecute source/etc/profile to make environment variables effective immediately5. Enter conf to copy the Zoo_sample.cfg file and name it as Zoo.cfg6. Vim zoo.cfg Edit configuration fi

Let's show the final results: http://gamebuilder.duapp.com/apprun.html?appid=genius-441422599398441Online editing interface:http://gamebuilder.duapp.com/gamebuilder.php?appid=genius-4414225993984411. Create a new project and delete the basketball and grass. Select tile vertically to place a new physical examination in the middle of the road dividing the road into 2. 2. Set the virtual height of the game scene to 5000,x direction and y-direction gravity set to 03. Place a counter to calculate th

. up, Switchspeed * Time. Deltatime);} else {//The menu is placed in a standard position after the rotation is complete currenttransform. Rotation= quaternion. Angleaxis(- -, Vector3. up);Nexttransform. Rotation= quaternion. Angleaxis(- -, Vector3. up);Set flag ready to rotate the next menu out Switchtag =1;}} if (Switchtag = =1) {if (nexttransform. Eulerangles. Y0) {Nexttransform. Rotate(Vector3. up, Switchspeed * Time. Deltatime);} else {Nexttransform. Rotation. Eulerangles. Set(0,0,0);Next me

1. I read a tutorial case of a marquee, and after a while I wrote a simple marquee. The problems encountered in the process are hereby recorded.The code is as follows:"Content-type"Content="text/html;charset=gb2312"> 'Button'Value='Stop Titleloop'Id='Stoploop'/>Problem:When I want to end the loop, the following code is not used to end the process:$ (' #stoploop '). Click (function() {Clearinterval (SetInterval ("Titleloop ()", 500))View clearinterval Description: The parameter of the Clearinterv

/* Define a Merry animation */@-webkit-keyframes marquee { 0% {left:0;} 100% {left:-100%;}} @keyframes Marquee { 0%{left:0} 100%{left: -100%}}.marquee { height:30px; Overflow:hidden; Position:relative;}. Marquee Div { display:block; width:200%; height:30px; Position:absolute; Overflow:hidden; -webkit-animation:marquee 4s linear infinite; Animation:marquee 4s linear Infinite;}[CSS3] Racing lights

Topic PortalTest instructions: The Chinese problem faceAnalysis: Put the official solution, that is, from 1 for the root node deep search record node depth, select the maximum depth of the point, will arrive at the point of the node are vis off, and then recalculate the depth of the point without vis, find the largest sum is the answer. Put a picture to understand:Harvest: The depth of the node of the compute treeCode:/************************************************* author:running_time* Create

Topic links Time limit: 20000ms single point time limit: 1000ms memory limit: 256MB description Fantasy township has a racetrack. There are n locations in the racetrack. There is also a one-way road between locations. These roads make the racetrack a structure of an outward tree. In other words, the road connects the n sites into a tree with a root. And all the edges are from the father pointing to the child. Because the fragrance likes to stimulate, every time she goes to the circuit will go f

Test instructions: Given a string of characters, u means uphill, d means downhill, F means flat, each has a different time to spend, ask you to walk from the beginning, the farthest can go.Analysis: Direct simulation is good, nothing to say, is to write down the time to remember double, because to return.The code is as follows:#include POJ 3672 Long Distance Racing (analog)

Test instructions: Tian Bogey horse race, ask you tin bogey most can win how much silver.Analysis: greedy, absolute greedy problem, greedy strategy is:1. If Tian bogey the fastest horse can catch up with the king, then directly win a game2. If the current slowest horse can catch up with the king, then win a game directly3. If Tian bogey current slowest horse can not exceed Qi Wang, then lose a game, and put King the fastest to killThrough the above strategy, is Tian bogey win the most.The code i

{static final int finish_line = 75;Private listPrivate Executorservice exec = Executors.newcachedthreadpool ();Private Cyclicbarrier barrier;Public horserace (int nhorses,final int pause) {Barrier = new Cyclicbarrier (nhorses,new Runnable () {@Overridepublic void Run () {StringBuilder s = new StringBuilder ();for (int i = 0; i S.append ("=");}System.out.println (s);for (Horse horse:horses) {System.out.println (Horse.tracks ());}for (Horse horse:horses) {if (Horse.getstrides () >= finish_line) {

Lufylegend engine is one of the simpler engines in the canvas game, it does not need to configure the environment, similar to the introduction of jquery package, reference to the corresponding JS file can beLufylegend Official website: http://www.lufylegend.com/First look at the gameGame design is rough, logic is also very simple, is to control the red car by clicking around to avoid the opposite of the car, if there is a collision, the game is over, and the game will be faster and faster.Game D

Topic Connection: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1333Test instructions: give you an n vertex, M edge (each edge has three parameters, opening time, closing time, and the time of passing this edge) of the direction graph;You need to find the shortest path from S to T; Dijkstra algorithm can be solved;Pit point: I use the queue optimization +vector store each edge, after each call Dijkstra, must initialize the adjacency table, in this place for a long time, this is the second, af

} * $ voidDijkstraints) {Panax Notoginseng for(intI=1; iINF; -d[s]=0; thememset (Used,0,sizeof(used)); + A for(intk=1; k){ the intu,m=INF; + for(intI=1; iif(!used[i]d[i]i]; -used[u]=1; $ for(inti=first[u];i!=-1; i=Next[i]) { $ intv=e[i].v,a=e[i].a,b=e[i].b,t=e[i].t; - - if(aContinue;//The open time is less than the time passed the - intnow=d[u]% (A +b);Wuyi if(now+t//now able to p