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Ultraviolet A 1428-Ping pong (tree array) Link to the question: Ultraviolet A 1428-Ping pong There are n table tennis fans living in a street and they often organize competitions. Each person has a different ability value. Each game requires three people. The referee must live between two players and the ability value must also be between players, ask how many matches can be held at most. Solution: Pre-proc

51Nod 1428 Event Scheduling issueslink:http://www.51nod.com/onlinejudge/questioncode.html#!problemid=14281428 scheduling issues Base time limit: 1 seconds space limit: 131072 KB Score: 10 Difficulty: 2 level algorithm problem has a number of activities, the first start and end time is [Si,fi], the same classroom arrangement of activities can not overlap, to arrange all activities, the minimum number of clas

1428 Event Scheduling issuesStart by starting from small to large sort first.In fact, as long as the minimum heap to maintain an end time, each time compared to the start time and the minimum time in the heap size, if larger than it is placed in the heap and the time will become the current task end time,Otherwise, we will open a new classroom. and add the end time to the heap, and be careful to determine if the heap is empty.#include #include#include

Link to the question: Ultraviolet A 1428-ping pong There are n table tennis fans living in a street and they often organize competitions. Each person has a different ability value. Each game requires three people. The referee must live between two players and the ability value must also be between players, ask how many matches can be held at most. Solution: Pre-processing Bi and CI are represented in 1 ~ In

;intN,Dd[4][2]={1,0,-1,0,0,1,0,-1},Map[ -][ -];//map: This map of the inputintVis[ -][ -],Map[ -][ -];//map: Shortest path map from (n-1,n-1) to all points__int64S[ -][ -],Num[ -][ -];structNode{intX,Y;}Ss;QueueNode>Q,Qq;voidBFs()//Production (n-1,n-1) to all points of the s

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 1428 I read the exercise question first. I thought it was graph theory and asked xsy to check the question .. In the end, you cannot give up. It was only after the end of the afternoon that it was originally a DP. Take the location of three vehicles as the status, J, K as the location of the car that is not moving, and I as the location of the current mobile car. DP [j, k, I] can be compo

Shortest Way +DPThe shortest path from (n,n) to each lattice is preprocessed first.Then you sort each point distance from large to small, and then you can recursively push the linear DP.#include #include#includestring>#include#include#include#includeusing namespacestd;Const intinf=0x7FFFFFFF;Const intmaxn= -;intN;intA[MAXN][MAXN],DIS[MAXN][MAXN];intdir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};structnode{intb; Node (intXintY) {a=x,b=y;}};structx{intX,y,dis;}

) is output for each group of test data ). Sample Input 31 2 31 2 31 2 331 1 11 1 11 1 Sample output 16 *************************************** **************************************** **************************************** * ****** question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1428 the q

Hdoj 1428 A Walk Through the Forest [spfa] + [memory-based search], hdojspfa A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission (s): 6397 Accepted Submission (s): 2348 Problem DescriptionJimmy experiences a lot of stress at work these days, especially since his accident made working difficult. to relax after a hard day, he likes to walk home. to make things even nicer, his office

Problem Description:LL recently addicted to AC extricate oneself, every day bedroom, room 2.1 line. Lack of exercise because of sitting on the computer for a long time. He decided to take full advantage of every time from the bedroom to the computer room to take a walk on campus. The entire HDU campus is a square layout that can be divided into n*n squares, representing each area. For example, I will live in the 18th dormitory located in the northwest corner of the campus, that is, the square (a

wants to know now is how many lines there are to meet the requirements. Can you tell him?Input The first behavior of each group of test data N (2=Output for each set of test data, the total number of routes is output (less than 2^63).Sample Input31 2 31 2 31 2 331 1 11 1 11 1 1Sample Output16 Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1428Title Analys

Title Address: HDU 1428First use the bfs+ priority queue to find the shortest distance from all points to the computer room, and then use the memory search.The code is as follows:#include #include #include #include #include #include #include #include #include using namespace STD;#define LL __int64#define PI ACOs ( -1.0)Const intMod=1e9+7;Const intinf=0x3f3f3f3f;Const Doubleeqs=1e-9;Const intmaxn= -+Ten;intD[MAXN][MAXN], MP[MAXN][MAXN], VIS[MAXN][MAXN], N; LL DP[MAXN][MAXN];intjx[]={0,0,1,-

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1428 Dijstra + dp; 1 #include View code

shortest distance from each point to the machine room is stored in a two-dimensional array dis[[];(3) We know that the rules of walking can only from K to K small, then if one of the routes, A, b two points, then a and a two point of the route direction is unique. So we can draw a theorem based on this point:the number of valid paths in a point to the room is equal to the sum of the number of points adjacent to it. By this time we have reduced the size of the search to a map that is O (n^2);Fou

UVA 1428-ping PongTopic linksTest instructions: Given some people, from left to right, each person has a skill value, now to hold the game, must meet the position from left to right 3 people. and the skill value from small to large or from big to small, ask a few forms of organizationIdea: Use a tree-like array to process the number of the left side of each position smaller than it and the number of small on the right. Then the left and right large ca

At the beginning of the submission, I was surprised to find out.Checked n times finally found strlen written in the For loop, so that the calculation of time complexity is n^2Problem Solving Ideas:Find the B string in the position where a string appears.Then we calculate the height array separately for a string, and then enumerate each position to find the substring that can be generated, and then the height can be reduced.#include Lightoj 1428 Melody

he wants to know now is how many lines there are to meet the requirements. Can you tell him? Input each set of test data of the first behavior N (2= Output outputs the total number of routes (less than 2^63) for each set of test data. Sample Input 3 1 2 3 1 2 3 1 2 3 3 1 1