说明：数据结构的扫描线算法，可以配合线段树等数据结构解决很多问题

说明：ACM practice question template

说明：HDU4022 ACM praction question

说明：HDU3864 ACM praction question

说明：HDU2586 ACM praction question

说明：HDU1711 ACM practice question

说明：常用的ACM算法模板，涉及常用的函数和STL，重要的公式和定理以及图论、几何等多种算法

说明：题目解答，可以ac，湖南大学acm算法题目

说明：基本功能实现，简单的算法设计，有助于ACM平台通过，当然这只是其中的一道题

说明：The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze. The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

说明：The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.

说明：Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

说明：Were Tian Ji lives in nowadays, he will certainly laugh at himself. Even more, were he sitting in the ACM contest right now, he may discover that the horse racing problem can be simply viewed as finding the maximum matching in a bipartite graph. Draw Tian's horses on one side, and the king's horses on the other. Whenever one of Tian's horses can beat one from the king, we draw an edge between them, meaning we wish to establish this pair. Then, the problem of winning as many rounds as possible is just to find the maximum matching in this graph. If there are ties, the problem becomes more complicated, he needs to assign weights 0, 1, or -1 to all the possible edges, and find a maximum weighted perfect matching...

说明：For example : There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4; One method : C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick. D could teach E;So D E are eligible to study on the same broomstick; Using this method , we need 2 broomsticks. Another method: D could teach A; So A D are eligible to study on the same broomstick. C could teach B; So B C are eligible to study on the same broomstick. E with no teacher or student are eligible to study on one broomstick. Using the method ,we need 3 broomsticks. ……

说明：diversity and capacity maximiazation

说明：acm是大学生程序竞赛中最常见的，对于一个程序员而言，参加acm不仅能够拓宽自己的知识面，而且能够在参加比赛的同时，认识一些新的朋友，而这些朋友可能在以后的生活中，对你产生很大的影响，对于一个非专业的程序员来说，如果能够参加acm竞赛，那么不仅能够培养自己的兴趣爱好，而且更能巩固知识面！

说明：acm入门参考资料，挑战程序设计竞赛pdf

说明：project euler代码库第五部分（已测试通过

说明：project euler代码库第四部分（已测试通过

说明：project euler代码库第三部分（已测试通过

说明：project euler代码库第二部分（已测试通过

说明：project euler代码库1,已测试通过

说明：这是ACM竞赛的试题库，由北京大学整理的。

说明：这是对ACM竞赛中对网络流和二分图的讲解。

说明：这些是我们学校培训的资料，都是ACM的经典考题。

说明：经典DP 背包问题，九讲背包入门，适合新手或入门者学习动态规划

说明：图书馆C++简单实现源码，功能较为简单，但具有基本增删改查，没有使用数据库，仅简单C++源代码，初学者使用

说明：ACM入门算法之dp，背包，高级数据结构，搜索，图，最短路。