uva 10032 Problem F: Tug of War】的更多相关文章

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=973 #include <cstdio> #include <cstring> #include <algorithm> #define ll long long using namespace std; ]; int n; ll dp[]; int mai…
紫皮书题: 题意:让你设计照明系统,给你n种灯泡,每种灯泡有所需电压,电源,每个灯泡的费用,以及每个灯泡所需的数量.每种灯泡所需的电源都是不同的,其中电压大的灯泡可以替换电压小的灯泡,要求求出最小费用 题解:每种电压灯泡要么全换,要么全不换,因为只换部分还要加额外的电源费用,并且换了部分之后费用更少,不如全换 先把灯泡按照电压从小到大排序,这样进行dp时,后面的电压大的如果比电压小的更优的话就可以替换了 设dp[j]为前j个最优了,则dp[i] = min{dp[i],dp[j] + (s[i]…
Description Problem F: Tug of War A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must n…
http://poj.org/problem?id=2576 二维数组01背包的变形. Tug of War Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8147   Accepted: 2191 Description A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must b…
Home Web Board ProblemSet Standing Status Statistics   Problem F: 求平均年龄 Problem F: 求平均年龄 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 720  Solved: 394[Submit][Status][Web Board] Description 定义一个Persons类,用于保存若干个人的姓名(string类型)和年龄(int类型),定义其方法 void ad…
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