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…

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…

题意:从 n个人里面找到m个人  每个人有两个值  d   p     满足在abs(sum(d)-sum(p)) 最小的前提下sum(d)+sum(p)最大 思路:dp[i][j]  i个人中  和是 j       运用背包的思想  二维背包 i是人数容量,人数要符合背包思想,每次只插入一个,逆序枚举 j是sum(d)+sum(p) 注意:这题的标准解法有误:https://blog.csdn.net/lyy289065406/article/details/6671105 这是有误的解法…

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…

题目: 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…

UVA323 Jury Compromise 题解 考虑用动态规划.该问题要求解的最终状态为,选出的 \(m\) 个人,使得辩方总分与控方总分差的绝对值最小,总分之和最大.即 \(\left| D(\mathcal{J}) - P(\mathcal{J}) \right|\) 最小,同时 \(D(\mathcal{J}) + P(\mathcal{J})\)最大. 使用三维 \(dp\) 数组记录状态,\(dp[i][j][k]\) 表示前 \(i\) 个人选出 \(j\) 个人同时辩方总分和为…

Jury Compromise Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:32355   Accepted:8722   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 p…

[题解]Jury Compromise(链表+DP) 传送门 题目大意 给你\(n\le 200\)个元素,一个元素有两个特征值,\(c_i\)和\(d_i\),\(c,d \in [0,20]\),现在请你选出\(m\le 20\)个元素使得\(\sum c+\sum d\)最大,使得$|\sum c - \sum d|最小,输出\sum c \(和\)\sum d$和一组合法方案. 分析 是DP无误了. 我们可以先不考虑绝对值,平移一下值域,假如说我们知道\(\sum c +- \sum d…

作为一个oier,以及大学acm党背包是必不可少的一部分.好久没做背包类动规了.久违地练习下-.- dd__engi的背包九讲:http://love-oriented.com/pack/ 鸣谢http://blog.csdn.net/eagle_or_snail/article/details/50987044,这里有大部分比较有趣的dp练手题. hud 2602 01背包板子题 #include<cstdio> #include<iostream> #include<cs…

K - Jury Compromise 参考:ACM POJ 1015 Jury Compromise(陪审团的人选,动态规划题,难) 说实话真有点难想,用一个DP[i][j]来表示在选取i个人,辩控差为j(j值已做些许处理)时辩控总分的最大值,用三个for循环来更新这个值.具体思路还是看参考博客吧.... 优先队列默认top()是最大值,如果写成priority_queue<int,vector<int>,greater<int> > top()则为最小值 代码: /…