http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=437 先介绍<algorithm>头文件中与集合运算有关的4个函数: set_union Union of two sorted ranges  (并集:A∪B) set_intersection Intersection of two sorted…

题目大意:给你两个集合,判断两个集合的关系(不相交.相等.真子集和其他).简单判断就可以了,不过STL的set没有交集.并集等操作有点让人觉得不方便... #include <cstdio> #include <iostream> #include <set> using namespace std; set<int> intersection(const set<int> &a, const set<int> &b)…

 A Puzzling Problem  The goal of this problem is to write a program which will take from 1 to 5 puzzle pieces such as those shown below and arrange them, if possible, to form a square. An example set of pieces is shown here. The pieces cannot be rota…

这道题要求我们求出图中的给定的两个节点(一个起点一个终点,但这是无向图)之间所有“路径中最大权值”的最小值,这无疑是动态规划. 我开始时想到根据起点和终点用动态规划直接求结果,但最终由于题中S过大,会超时. 超时的代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <string…

uva 10192 Vacation The Problem You are planning to take some rest and to go out on vacation, but you really don't know which cities you should visit. So, you ask your parents for help. Your mother says "My son, you MUST visit Paris, Madrid, Lisboa an…

题意:给出几个圆的半径,贴着底下排放在一个长方形里面,求出如何摆放能使长方形底下长度最短. 由于球的个数不会超过8, 所以用全排列一个一个计算底下的长度,然后记录最短就行了. 全排列用next_permutation函数,计算长度时坐标模拟着摆放就行了. 摆放时折腾了不久,一开始一个一个把圆放到最左端,然后和前面摆好的圆比较检查是否会出现两个圆重叠,是的话就把当前圆向右移动到相切的位置.然后判断宽度. 结果发现过不了几个例子,检查了好久,终于发现问题出在初始位置上,如果出现一个特别小的圆放到最左…

题目: Given a list of numbers that may has duplicate numbers, return all possible subsets Notice Each element in a subset must be in non-descending order. The ordering between two subsets is free. The solution set must not contain duplicate subsets. Ex…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2483 Hyper Prefix Sets Prefix goodness of a set string islength of longest common prefix*number of strings in the set.For example the prefix goodnes…

就是给出非常多点,要求分成两个集合,在同一个集合里的点要求随意两个之间的距离都大于5. 求一个集合.它的点数目是全部可能答案中最少的. 直接从随意一个点爆搜,把它范围内的点都丢到跟它不一样的集合里.不断这样搞即可了. 由于可能有非常多相离的远,把每次搜索得到的那个最小的数目加起来就可以. 因为全部点都格点上,所以仅仅须要枚举一个点可以包括的点是否在数据中存在就可以. 当然也能够用一棵树直接去找.这我并不会.. 时间复杂度是81nlogn 湖大的OJ机器太老.. .还要开栈. . .UVA LIV…

Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example,If nums = [1,2,2], a solution is: [ [2], [1], [1,2,2], [2,2], [1,2], [] ] 分析: 因为有重…