problem 1029. Two City Scheduling 参考 1. Leetcode_easy_1029. Two City Scheduling; 完…

题目如下: There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1]. Return the minimum cost to fly every person to a city such tha…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 小根堆 排序 日期 题目地址:https://leetcode.com/problems/two-city-scheduling/ 题目描述 There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is c…

[BZOJ1645][Usaco2007 Open]City Horizon 城市地平线 Description Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings. The entire ho…

题目链接:http://www.wikioi.com/problem/1029/ 算法:数学 本题有个2小技巧. 一棵二叉树的前序遍历a1a2a3...ai和后序遍历b1b2b3...bi有一种关系:当只有一棵子树的根 在a序列下标为i, 在b序列下标为b有  a[i-1] == b[j+1]这是因为当根只有一棵子树时,前序和后序遍历都是先遍历它的孩子,而且是唯一的一个孩子,所以相对位置是一样的. 当确定了一棵树的前序和后序遍历时,可以得到用如上方法找到只有一棵子树的根,并且因为上面的性质,子树…

http://www.lydsy.com/JudgeOnline/problem.php?id=1029 按右端点排序后依次加入,并且每一次看是否能被修筑,如果能就修:否则查找原来修过的,如果原来修过的最大的建筑花的时间比当前所要花的时间大,那么我们就决策:不修原来那个,改为修当前的(因为起点一样,所花时间少,两者的右边界都是满足的,用了后者时间能减少,使得给可能存在的后边的解提供机会).可以证明这样最优. #include <cstdio> #include <cstring>…

Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the medi…

problem 1021. Remove Outermost Parentheses 参考 1. Leetcode_easy_1021. Remove Outermost Parentheses; 完…

problem 1022. Sum of Root To Leaf Binary Numbers 参考 1. Leetcode_easy_1022. Sum of Root To Leaf Binary Numbers; 完…

problem 1025. Divisor Game 参考 1. Leetcode_easy_1025. Divisor Game; 完…