POJ 1015 Jury Compromise（双塔dp）

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 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

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

题意：

从n个人中选m个人，要求sum1-sum2最小，sum1+sum2最大.

思路：

来自于：http://exp-blog.com/2018/06/25/pid-1172/

现用dp(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dp(j, k)”）的辩控和。

可行方案dp(j-1, x)能演化成方案dp(j, k)的必要条件是：存在某个候选人i，i 在方案dp(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dp(j-1, x)中，选出 dp(j-1, x) + S(i) 的值最大的那个，那么方案dp(j-1, x)再加上候选人i，就演变成了方案 dp(j, k)。

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<set>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<ctime>

#define fuck(x) cout<<#x<<" = "<<x<<endl;

#define ls (t<<1)

#define rs ((t<<1)+1)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = ;

const int inf = 2.1e9;

const ll Inf = ;

const int mod = ;

const double eps = 1e-;

const double pi = acos(-);

int n,m;

int dp[][];

int num1[],num2[],sum[],diff[];

int path[][];

int fix = ;

bool check(int i,int k,int j){

    while(i&&path[i][k]!=j){

        k-=diff[path[i][k]];

        i--;

    }

    return i==;

}

int main()

{

//    ios::sync_with_stdio(false);

//    freopen("in.txt","r",stdin);

    int cases=;

    while(scanf("%d%d",&n,&m)!=EOF&&(n||m)){

        vector<int>ans;

        for(int i=;i<=n;i++){

            scanf("%d%d",&num1[i],&num2[i]);

            sum[i]=num1[i]+num2[i];

            diff[i]=num1[i]-num2[i];

        }

        memset(dp,-,sizeof(dp));

        dp[][fix]=;

        for(int i=;i<=m;i++){

            for(int k=;k<=*fix;k++){

                if(dp[i-][k]<){continue;}

                for(int j=;j<=n;j++){

                    if(dp[i][k+diff[j]]<dp[i-][k]+sum[j]&&check(i-,k,j)){

                        dp[i][k+diff[j]]=dp[i-][k]+sum[j];

                        path[i][k+diff[j]]=j;

                    }

                }

            }

        }

        int minn=,maxx=;

        for(int i=;i<=fix;i++){

//            cout<<dp[m][fix-i];

            if(dp[m][fix-i]>=||dp[m][fix+i]>=){

                minn=i;break;

            }

        }

        int rec=;

        if(dp[m][fix-minn]>dp[m][fix+minn]){

            rec=fix-minn;

            maxx=dp[m][fix-minn];

        }

        else{

            rec=fix+minn;

            maxx=dp[m][fix+minn];

        }

        int ans1,ans2;

        ans1=ans2=;

        while(m){

            ans.push_back(path[m][rec]);

            ans1+=num1[path[m][rec]];

            ans2+=num2[path[m][rec]];

            rec-=diff[path[m][rec]];

            m--;

        }

        cases++;

        printf("Jury #%d\n",cases);

        printf("Best jury has value %d for prosecution and value %d for defence:\n",ans1,ans2);

        sort(ans.begin(),ans.end());

        int sz=ans.size();

        for(int i=;i<sz;i++){

            printf(" %d",ans[i]);

        }

        printf("\n\n");

    }

    return ;

}

巴特西

POJ 1015 Jury Compromise（双塔dp）

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