http://poj.org/problem?id=1015   Description In Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general public. Every time a trial is set to begin, a jury has to be selected, which is do…

描述在遥远的国家佛罗布尼亚,嫌犯是否有罪,须由陪审团决定.陪审团是由法官从公众中挑选的.先随机挑选n个人作为陪审团的候选人,然后再从这n个人中选m人组成陪审团.选m人的办法是: 控方和辩方会根据对候选人的喜欢程度,给所有候选人打分,分值从0到20.为了公平起见,法官选出陪审团的原则是:选出的m个人,必须满足辩方总分和控方总分的差的绝对值最小.如果有多种选择方案的辩方总分和控方总分的之差的绝对值相同,那么选辩控双方总分之和最大的方案即可.输入输入包含多组数据.每组数据的第一行是两个整数n和m,n是…

http://acm.fzu.edu.cn/problem.php?pid=1005 Description The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurant and supplying several…

Jury Compromise Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33737   Accepted: 9109   Special Judge Description In Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general…

提议:在遥远的国家佛罗布尼亚,嫌犯是否有罪,须由陪审团决定.陪审团是由法官从公众中挑选的.先随机挑选n个人作为陪审团的候选人,然后再从这n个人中选m人组成陪审团.选m人的办法是:控方和辩方会根据对候选人的喜欢程度,给所有候选人打分,分值从0到20.为了公平起见,法官选出陪审团的原则是:选出的m个人,必须满足辩方总分和控方总分的差的绝对值最小.如果有多种选择方案的辩方总分和控方总分的之差的绝对值相同,那么选辩控双方总分之和最大的方案即可. 题解:开始想到的是二维01背包,因为评价差的总分值最大可能…

题目不难,暴力地dp一下就好,但是不知道我WA在哪里了,对拍了好多的数据都没找出错误= =.估计又是哪里小细节写错了QAQ..思路是用dp[i][j]表示已经选了i个,差值为j的最大和.转移的话暴力枚举当前选那个即可.代码如下(WA的,以后有机会再找找错在哪里吧0.0): #include <stdio.h> #include <algorithm> #include <string.h> #include <set> using namespace std…

In Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general public. Every time a trial is set to begin, a jury has to be selected, which is done as follows. First, several people are draw…

\(Jury Compromise\) \(solution:\) 这道题很有意思,它的状态设得很...奇怪.但是它的数据范围实在是太暴露了.虽然当时还是想了好久好久,出题人设了几个限制(首先要两个的总和差值最小)(然后需要让它的总和最大).我们发现每一个人的顺序是无关紧要的,这其实又提示了我们可以背包.但我们发现很难设状态,我们需要让我们的总差值接近0,但是我们在加人的时候我们的总差值可能会增大也可能会减小,这不符合背包的基本要求,所以我们不能将总差值设为我们背包的权值.于是一个奇妙的想法产生…

Charm Bracelet    POJ 3624 就是一道典型的01背包问题: #include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; ],b[]; ]; int main() { int n,m,i,j; while(scanf("%d%d",&n,&m)!=EOF) { ;i&l…

Scout YYF I Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5565   Accepted: 1553 Description YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties,…