POJ 1015

Jury Compromise

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23258 Accepted: 6012 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 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

参照网上一位大神的思路：

题目大意
这是一道很不错的动态规划DP题，题目大意是有n组数据，每组数据有两个属性，从中选取m个数据，使得属性一之和和属性二之和的差最小，如果同时存在多组差相同的，则选择属性一之和和属性二之和的和最大的。同时要记录选取的数据集。
解题思路
这是一道很好的动态规划题目，要求记录中间的过程。我用了两种方法做了下。
第一种方法：
dp[i][j][k]表示前i个数据中选取j个数据两种属性差为k时的两种属性和。
dp[i][j][k]=max(dp[i-1][j][k],dp[i-1][j-1][k-d[i]]+p[i])
这样用path[i][j][k]表示对应的轨迹，如果使用了第i个数据，则path[i][j][k]=I,这样前一个数据就是path[i-1][j-1][k-d[i]],以此类推下去就可以得到全部的轨迹。
第二种方法：
dp[j][k]表示选取j个数据属性差为k时的属性和
dp[j][k]=max(dp[j-1][k-d[i]]),其中i=1,2,3,…n
这个时候可以用path[j][k]=i表示对应的路径，path[j][k]的上一个路径就是path[j-1][k-d[path[j][k]]]
 无论哪一种方法，由于差值可以为负，所以DP的差值状态要有一个便宜，20个数据最大为20，总共就是400，所以可以用400做偏移量，很多人都是这样做的，但是我的却出问题了，因为中间有一处差值从0到800时0减去了一个正的差值，会导致结果为负，所以我就用了420，这点应该是可以改进的。

第一种思路：

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)const int MAXN = 100000 + 50;const int maxw = 100 + 20;const int MAXNNODE = 1000000 +10;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;#define eps 1e-8#define mod 1000000007typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pi;int dp[210][25][900];int diff[210] , rst[25];int path[210][25][900];int main(){    //ios::sync_with_stdio(false);#ifdef Online_Judge    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif // Online_Judge    int n ,  m , a , b , k , len1 , len2,  p  , q , Case = 0;    while(scanf("%d%d", &n , &m) , (n||m))    {        printf("Jury #%d\n" , ++Case);        len1 = 420 - 20 * m,len2 = 420 + 20 * m;        clr(path , 0);        clr(dp , -1);        clr(diff , 0);        scanf("%d%d" , &a , &b);        diff[1] = a - b;        int temp = 420 + a - b;        dp[1][1][temp] = a + b;        path[1][1][temp] = 1;        FORR(i , 2 , n)        {            scanf("%d%d" , &a , &b);            temp = 420 + a - b;            diff[i] = a - b;            FORR(k , len1 , len2)dp[i][1][k] = dp[i - 1][1][k];            if(a + b > dp[i][1][temp])            {                path[i][1][temp] = i;                dp[i][1][temp] = a + b;            }            for(int j  = 2 ; j <= ( m > i ? i : m);j++)            {                FORR(k , len1, len2)                {                    p = dp[i - 1][j][k] , q = dp[i - 1][j - 1][k - a + b];                    if(q == -1)                    {                        dp[i][j][k] = p;                    }                    else                    {                        if(p >= q + a + b)dp[i][j][k] = p;                        else {dp[i][j][k] = q + a + b ; path[i][j][k] = i;}                    }                }            }        }        int i = 0;        int ll , pro , def;        while(dp[n][m][420 + i] == -1&&dp[n][m][420 - i] == -1)i++;        if(dp[n][m][420 + i] > dp[n][m][420 - i])        {            pro = (dp[n][m][420 + i] + i) >> 1;            def = (dp[n][m][420 + i] - i) >> 1;            ll = 420 + i;        }        else        {            pro = (dp[n][m][420 - i] - i) >> 1;            def = (dp[n][m][420 - i] + i) >> 1;            ll = 420 - i;        }        int j = m;        while(j)        {            a = path[n][j][ll];            if(a)            {                rst[j] = a;                ll -= diff[a];                j--;            }            n--;        }        printf("Best jury has value %d for prosecution and value %d for defence:\n" , pro , def);        FORR(i , 1 ,m)printf(" %d",rst[i]);        printf("\n");    }    return 0;}

第二种思路：

#include"iostream"#include"algorithm"using namespace std;int d[201],p[201],dp[21][841],path[21][841];bool isvisited(int i,int j,int k){int a; while(i){a=path[i][j]; if(k==a)return true; i--;j-=p[a]; } return false; } int main(){int i,j,k,a,b,len1,len2,m,n,index=0;while(cin>>n>>m&&n!=0&&m!=0)    {index++; cout<<"Jury #"<<index<<endl; memset(dp,-1,sizeof(dp));memset(path,0,sizeof(path)); len1=420-20*m;len2=420+20*m; for(i=1;i<=n;i++){ scanf("%d %d",&a,&b);d[i]=a+b;p[i]=a-b; if(dp[1][a-b+420]<a+b){dp[1][a-b+420]=a+b;path[1][a-b+420]=i;} }for(i=2;i<=m;i++){for(j=len1;j<=len2;j++){for(k=1;k<=n;k++) { if(dp[i-1][j-p[k]]!=-1&&dp[i][j]<dp[i-1][j-p[k]]+d[k]&&!isvisited(i-1,j-p[k],k)){dp[i][j]=dp[i-1][j-p[k]]+d[k];path[i][j]=k;  } } } }int ll,rst[21],pro,def; i=0; while(dp[m][420+i]==-1&&dp[m][420-i]==-1)i++; if(dp[m][420+i]>dp[m][420-i]){pro=(dp[m][420+i]+i)/2;def=(dp[m][420+i]-i)/2;ll=420+i; } else{pro=(dp[m][420-i]-i)/2;def=(dp[m][420-i]+i)/2;ll=420-i; }         j=m;         while(j){a=path[j][ll];rst[j]=a;             j--;ll-=p[a];         }        sort(rst+1,rst+m+1);         cout<<"Best jury has value "<<pro<<" for prosecution and value "<<def<<" for defence:"<<endl;         for(i=1;i<=m;i++)printf(" %d",rst[i]);cout<<endl<<endl; } }