Building roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7019   Accepted: 2387 Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some ro…

POJ 2594 Treasure Exploration 题目链接 题意:有向无环图,求最少多少条路径能够覆盖整个图,点能够反复走 思路:和普通的最小路径覆盖不同的是,点能够反复走,那么事实上仅仅要在多一步.利用floyd求出传递闭包.然后依据这个新的图去做最小路径覆盖就可以 代码: #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using names…

HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链接HDU 题目链接POJ 题意: 有n个牛棚, 还有两个中转站S1和S2, S1和S2用一条路连接起来. 为了使得随意牛棚两个都能够有道路联通,如今要让每一个牛棚都连接一条路到S1或者S2. 有a对牛棚互相有仇恨,所以不能让他们的路连接到同一个中转站. 还有b对牛棚互相喜欢,所以他们的路必须连到同一个中专站.…

POJ 3126 Prime Path(素数路径) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on…

POJ 2749 分解因数(计蒜客) Description 给出一个正整数a,要求分解成若干个正整数的乘积,即a = a1 * a2 * a3 * - * an,并且1 < a1 <= a2 <= a3 <= - <= an,问这样的分解的种数有多少.注意到a = a也是一种分解. Input 第1行是测试数据的组数n,后面跟着n行输入.每组测试数据占1行,包括一个正整数a (1 < a < 32768) Output n行,每行输出对应一个输入.输出应是一个正…

Building roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6091   Accepted: 2046 Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some ro…

http://poj.org/problem?id=3020 Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7565   Accepted: 3758 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile ph…

http://poj.org/problem?id=2060 Taxi Cab Scheme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5459   Accepted: 2286 Description Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination…

POJ 3216 Repairing Company id=3216">题目链接 题意:有m项任务,每项任务的起始时间,持续时间,和它所在的block已知,且往返每对相邻block之间的时间也知道,问最少须要多少个工人才干完毕任务,即x最少是多少 思路:先floyd求出每两个block之间的最小距离,然后就是最小路径覆盖问题,一个任务之后能赶到还有一个任务就建边 代码: #include <cstdio> #include <cstring> #include &l…

//yy:昨天看着这题突然有点懵,不知道怎么记录路径,然后交给房教了,,,然后默默去写另一个bfs,想清楚思路后花了半小时写了120+行的代码然后出现奇葩的CE,看完FAQ改了之后又WA了.然后第一次用对拍去找特殊数据折腾到十二点半终于AC,最后只想感叹没有仔细读题,没办法啊看着英语略烦躁,不扯了,那个题题解不想写了,回到这题...今天中午还是花了四十分钟写了这题(好慢啊orz),感觉,啊为什么别人可以一下子就写出来,我却想不到怎么写呢!!! poj 3414 Pots  [BFS] 题意:两个…