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[dp] poj 1015 Jury Compromise
题目链接：
http://poj.org/problem?id=1015
Jury Compromise
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 done as follows. First, several people are drawn randomly from the public. For each person in this pool, defence and prosecution assign a grade from 0 to 20 indicating their preference for this person. 0 means total dislike, 20 on the other hand 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 order to ensure a fair trial, the tendencies of the jury to favour either defence or prosecution should be as balanced as possible. The jury therefore has to be chosen in a way that is satisfactory to both parties. We will now make this more precise: given a pool of n potential jurors and two values di (the defence‘s value) and pi (the prosecution‘s value) for each potential juror i, you are to select a jury of m persons. If J is a subset of {1,..., n} with m elements, then D(J ) = sum(dk) k belong to J and P(J) = sum(pk) k belong to J are the total values of this jury for defence and prosecution. For an optimal jury J , the value |D(J) - P(J)| must be minimal. If there are several 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 both parties. You are to write a program that implements this jury selection process and chooses an optimal jury given a set of candidates. Input The input file contains several jury selection rounds. Each round starts with a line containing two integers n and m. n is the number of candidates and m the number of jury members. These values will satisfy 1<=n<=200, 1<=m<=20 and of course m<=n. The following n lines contain the two integers pi and di for i = 1,...,n. A blank line separates each round from the next. The file ends with a round that has n = m = 0. Output For each round output a line containing the number of the jury selection round (‘Jury #1‘, ‘Jury #2‘, etc.). On the next line print the values D(J ) and P (J ) of your jury as shown below and on another line print the numbers of the m chosen candidates in ascending order. Output a blank before each individual candidate number. Output an empty line after each test case. Sample Input 4 2 1 2 2 3 4 1 6 2 0 0 Sample Output Jury #1 Best jury has value 6 for prosecution and value 4 for defence: 2 3 Hint If your solution is based on an inefficient algorithm, it may not execute in the allotted time. Source Southwestern European Regional Contest 1996 | ||||||||||
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题目意思：
有n个人，每个人有个d值和p值，求选出m个人，使得|sum(d)-sum(p)|最小，如果最小值相同，则选择|sum(d)+sum(p)|较大的。输出选出的人。
n<=200 d,p<20 m<=20
解题思路：
dp,实际上就是一个背包。
|sum(d)-sum(p)|最小，实际上就是找能凑成的最靠近0的.由于-20=<d-p<=20,所以把每个d-p都加上20，变成非负的整数。然后算出最靠近20*m的就是满足要求的最小的。
dp[i][j]：表示当sum(d)-sum(p)为i，且有j个人时，能否凑成。保存一个add[i][j]：表示这j个人的sum(d)+sum(p)
最后在20*m附近找.
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 8100
int dp[Maxn][25],add[Maxn][25];
int savea[220],saveb[220];
vector<int>re[Maxn][25];
int n,m;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int ca=0;
while(scanf("%d%d",&n,&m)&&n+m)
{
for(int i=1;i<=n;i++)
scanf("%d%d",&savea[i],&saveb[i]);
memset(dp,0,sizeof(dp));
memset(add,0,sizeof(add));
dp[0][0]=1;
int lim=20*n*2;
for(int i=0;i<=lim;i++)
for(int j=0;j<=m;j++)
re[i][j].clear();
for(int i=1;i<=n;i++)
{
int temp=savea[i]-saveb[i]+20;
for(int j=lim;j>=temp;j--)
{
for(int k=m;k>=1;k--)
{
if(dp[j-temp][k-1])
{
int te=add[j-temp][k-1]+savea[i]+saveb[i];
if(!dp[j][k])
{
add[j][k]=te;
re[j][k]=re[j-temp][k-1];
re[j][k].push_back(i);
dp[j][k]=1;
}
else if(te>add[j][k])
{
add[j][k]=te;
re[j][k]=re[j-temp][k-1];
re[j][k].push_back(i);
}
}
}
}
}
int ansa,ansb,a;
int p=0,pp=20*m;
int k=m;
while(!dp[pp+p][k]&&!dp[pp-p][k])
p++;
if(!dp[pp+p][k])
{
ansa=abs((pp-p+add[pp-p][k]-20*m))/2;
ansb=(add[pp-p][k]-pp+p+20*m)/2;
a=pp-p;
//printf("Best jury has value %d for prosecution and value %d for defence:\n",ansa,)
}
else if(!dp[pp-p][k])
{
ansa=abs((pp+p+add[pp+p][k]-20*m))/2;
ansb=(add[pp+p][k]-pp-p+20*m)/2;
a=pp+p;
}
else
{
if(add[pp+p][k]>add[pp-p][k])
{
ansa=abs((pp+p+add[pp+p][k]-20*m))/2;
ansb=(add[pp+p][k]-pp-p+20*m)/2;
a=pp+p;
}
else
{
ansa=abs((pp-p+add[pp-p][k]-20*m))/2;
ansb=(add[pp-p][k]-pp+p+20*m)/2;
a=pp-p;
}
}
printf("Jury #%d\n",++ca);
printf("Best jury has value %d for prosecution and value %d for defence:\n",ansa,ansb);
for(int i=0;i<re[a][k].size();i++)
printf(" %d",re[a][k][i]);
printf("\n\n");
}
return 0;
}
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