Mar 14, 2022 陪审团人选 DescriptionIn Frobnia, a far-away country, the verdicts in court trials are determinedby a jury consisting of members of the general public. Every time a trial isset to begin, a jury has to be selected, which is done as follows. First,several people are drawn randomly from the public. For each person inthis pool, defence and prosecution assign a grade from 0 to 20 indicatingtheir preference for this person. 0 means total dislike, 20 on the otherhand means that this person is considered ideally suited for the jury.Based on the grades of the two parties, the judge selects the jury. In orderto ensure a fair trial, the tendencies of the jury to favour either defence orprosecution should be as balanced as possible. The jury therefore has tobe chosen in a way that is satisfactory to both parties.We will now make this more precise: given a pool of n potential jurorsand two values di (the defence’s value) and pi (the prosecution’svalue) for each potential juror i, you are to select a jury of m persons. If Jis a subset of {1,…, n} with m elements, then D(J ) = sum(dk) k belong to Jand P(J) = sum(pk) k belong to J are the total values of this jury fordefence and prosecution.For an optimal jury J , the value |D(J) - P(J)| must be minimal. If there areseveral jurys with minimal |D(J) - P(J)|, one which maximizes D(J) + P(J)should be selected since the jury should be as ideal as possible for bothparties.You are to write a program that implements this jury selection processand chooses an optimal jury given a set of candidates.InputThe input file contains several jury selection rounds. Each round startswith a line containing two integers n and m. n is the number ofcandidates and m the number of jury members.These values will satisfy 1<=n<=200, 1<=m<=20 and of course m<=n. Thefollowing n lines contain the two integers pi and di for i = 1,…,n. A blankline separates each round from the next.The file ends with a round that has n = m = 0.OutputFor each round output a line containing the number of the jury selectionround (‘Jury #1’, ‘Jury #2’, etc.).On the next line print the values D(J ) and P (J ) of your jury as shownbelow and on another line print the numbers of the m chosen candidatesin ascending order. Output a blank before each individual candidatenumber.Output an empty line after each test case.Sample Input“4 21 22 34 16 20 0”Sample Output“Jury #1”Best jury has value 6 for prosecution and value 4 for defence: 2 3HintIf your solution is based on an inefficient algorithm, it may not execute inthe allotted time.SourceSouthwestern European Regional Contest 1996中文：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是候选人数目，m是陪审团人数。注意，1<=𝑛<=200,1<=𝑚<=20,𝑚<=𝑛。接下来的n行，每行表示一个候选人的信息，它包含2个整数，先后是控方和辩方对该候选人的打分。候选人按出现的先后从1开始编号。两组有效数据之间以空行分隔。最后一组数据𝑛=𝑚=0输出对每组数据，先输出一行，表示答案所属的组号,如 ‘Jury #1’, ‘Jury#2’, 等。接下来的一行要象例子那样输出陪审团的控方总分和辩方总分。再下来一行要以升序输出陪审团里每个成员的编号，两个成员编号之间用空格分隔。每组输出数据须以一个空行结束。样例输入“4 21 22 34 16 20 0”样例输出“Jury #1”Best jury has value 6 for prosecution and value 4 for defence: 2 3为叙述问题方便，现将任一选择方案中，辩方总分和控方总分之差简称为“辩控差”，辩方总分和控方总分之和称为“辩控和”。第i个候选人的辩方总分和控方总分之差记为𝑉(𝑖)，辩方总分和控方总分之和记为𝑆(𝑖)。现用𝑓(𝑗,𝑘)表示，取j个候选人，使其辩控差为k的所有方案中，辩控和最大的那个方案（该方案称为“方案𝑓(𝑗,𝑘)”）的辩控和。并且，我们还规定，如果没法选j个人，使其辩控差为k，那么𝑓(𝑗,𝑘)的值就为-1，也称方案𝑓(𝑗,𝑘)不可行。本题是要求选出m个人，那么，如果对k的所有可能的取值，求出了所有的𝑓(𝑚,𝑘)(−20×𝑚≤𝑘≤20×𝑚)，那么陪审团方案自然就很容易找到了。问题的关键是建立递推关系。需要从哪些已知条件出发，才能求出𝑓(𝑗,𝑘)呢？显然，方案𝑓(𝑗,𝑘)是由某个可行的方案𝑓(𝑗−1,𝑥)(−20×𝑚≤𝑥≤20×𝑚)演化而来的。可行方案𝑓(𝑗−1,𝑥)能演化成方案𝑓(𝑗,𝑘)的必要条件是：存在某个候选人i，i在方案𝑓(𝑗−1,𝑥)中没有被选上，且𝑥+𝑉(𝑖)=𝑘。在所有满足该必要条件的𝑓(𝑗−1,𝑥)中，选出𝑓(𝑗−1,𝑥)+𝑆(𝑖)的值最大的那个，那么方案𝑓(𝑗−1,𝑥)再加上候选人i，就演变成了方案𝑓(𝑗,𝑘)。这中间需要将一个方案都选了哪些人都记录下来。不妨将方案𝑓(𝑗,𝑘)中最后选的那个候选人的编号，记在二维数组的元素𝑝𝑎𝑡ℎ[𝑗][𝑘]中。那么方案𝑓(𝑗,𝑘)的倒数第二个人选的编号，就是𝑝𝑎𝑡ℎ[𝑗−1][𝑘−𝑉[𝑝𝑎𝑡ℎ[𝑗][𝑘]]。假定最后算出了解方案的辩控差是k，那么从𝑝𝑎𝑡ℎ[𝑚][𝑘]出发，就能顺藤摸瓜一步步求出所有被选中的候选人。初始条件，只能确定𝑓(0,0)=0。由此出发，一步步自底向上递推，就能求出所有的可行方案𝑓(𝑚,𝑘)(−20×𝑚≤𝑘≤20×𝑚)。实际解题的时候，会用一个二维数组f来存放𝑓(𝑗,𝑘)的值。而且，由于题目中辩控差的值k可以为负数，而程序中数租下标不能为负数，所以，在程序中不妨将辩控差的值都加上400，以免下标为负数导致出错，即题目描述中，如果辩控差为0，则在程序中辩控差为400。1#include<stdio.h>2#include<stdlib.h>3#include<iostream>4#include<string.h>5usingnamespace std;6int f[30][1000];7//f[j,k]表示：取j个候选人，使其辩控差为k的方案中8//辩控和最大的那个方案（该方案称为“方案f(j,k)”)的控辩和9int Path[30][1000];10//Path数组用来记录选了哪些人11//方案f(j,k)中最后选的那个候选人的编号，记在Path[j][k]中12int P[300];//控方打分13int D[300]; //辩方打分1415int Answer[30];//存放最终方案的人选1617int cmp(constvoid*a,constvoid*b)18{19 return*(int*)a-*(int*)b;20}2122int main()23{24 int i,j,k;25 int t1,t2;26 int n,m;27 int nMinP_D;//辩控双方总分一样时的辩控差28 int iCase;//测试数据编号29 iCase=0;30 while(scanf("%d %d",&n,&m))31 {32 if(n==0&&m==0)break;33 iCase++;34 for(i=1;i<=n;i++)35 scanf("%d %d",&P[i],&D[i]);36 memset(f,-1,sizeof(f));37 memset(Path,0,sizeof(Path));38 nMinP_D=m*20;//题目中的辩控差为0，对应于程序中的辩控差为m*2039 f[0][nMinP_D]=0;40 for(j=0;j<m;j++)//每次循环选出第j个人，共要选出m人41 {42 for(k=0;k<=nMinP_D*2;k++)//可能的辩控差为[0，nMinP_D*2]43 if(f[j][k]>=0)//方案f[j,k]可行44 {45 for(i=1;i<=n;i++)46 if(f[j][k]+P[i]+D[i]>f[j+1][k+P[i]-D[i]])47 {48 t1=j;t2=k;49 while(t1>0&&Path[t1][t2]!=i)//验证i是否在前面出现过50 {51 t2-=P[Path[t1][t2]]-D[Path[t1][t2]];52 t1--;53 }54 if(t1==0)55 {56 f[j+1][k+P[i]-D[i]]=f[j][k]+P[i]+D[i];57 Path[j+1][k+P[i]-D[i]]=i;58 }59 }60 }61 }62 i=nMinP_D;63 j=0;64 while(f[m][i+j]<0&&f[m][i-j]<0) j++;65 if(f[m][i+j]>f[m][i-j]) k=i+j;66 else k=i-j;67 printf("Jury #%d\n",iCase);68 printf("Best jury has value %d for prosecution and value%d for defence:\n",(k-nMinP_D+f[m][k])/2,(f[m][k]-k+nMinP_D)/2);69 for(i=1;i<=m;i++)70 {71 Answer[i]=Path[m-i+1][k];72 k-=P[Answer[i]]-D[Answer[i]];73 }74 qsort(Answer+1,m,sizeof(int),cmp);75 for(i=1;i<=m;i++)76 printf(" %d",Answer[i]);77 printf("\n\n");7879 }80 return 0;81}