SGU 280.Trade centers 解题报告 题意: n(<=30000)个城市,(n-1)条道路,求最少需要选择多少个城市建造市场,使得所有城市到任意一个市场的距离不大于k. Solution: 比较好的贪心题.实现起来也有一定技巧. 先以任意点为根,构造出一颗有根树. 首先比较容易想到的是从叶子节点向上寻找,如果只有一个距离为k的点,就把它选上.但是有多个呢? 于是思考更一般的做法,由于是树形结构,先考虑以x节点为根的子树.我们先假设f[x]代表离x节点向下的市场点的距离,这个值可以…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44470    Accepted Submission(s): 14872 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pound…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 36632    Accepted Submission(s): 12064 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats gu…

题意:就是老鼠要用猫粮换粮食,第i个房间一些东西,要用东西去换,可以不全换.问给定的猫粮最多能换多少粮食. 析:贪心算法.我们先算出来每个房间物品的平均价格是多少,肯定越低越好,并且如果能全换就全换,如果不能, 肯定是最后一次了,就把剩下全部换了,看看能换多少.求和. 代码如下: #include <iostream> #include <cstdio> #include <algorithm> #include <queue> #include <v…

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains Jii pounds of JavaBeans and requires Fii pounds of cat food. FatMous…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 60654    Accepted Submission(s): 20406 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds…

解题思路:一只老鼠共有m的猫粮,给出n个房间,每一间房间可以用f[i]的猫粮换取w[i]的豆,问老鼠最多能够获得豆的数量 sum 即每一间房间的豆的单价为v[i]=f[i]/w[i],要想买到最多的豆,一定是先买最便宜的,再买第二便宜的,再买第三便宜的 -----m的值为0的时候求得的sum即为最大值   所以先将v[i]从小到大排序.   FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1009 题目大意:肥鼠准备了 磅的猫粮,准备和看管仓库的猫交易,仓库里装有他最喜爱的食物 豆.仓库有 个房间.第 间房包含了 磅的 豆,需要 磅的猫粮.肥鼠不必为了房间中的所有 豆而交易,相反,他可以支付 % 磅的猫粮去交换得到 % 磅的 豆.这里, 表示一个实数.现在他将这项任务分配给了你:请告诉他,能够获得的 豆的最大值是多少. 题目要求:(输入)输入包含多组测试数据.对于每组测试数据,以包含了两…