[kuangbin 基础dp][POJ 1015] Jury Compromise（dp）

Posted 2021-03-10 mizersy

[kuangbin 基础dp][POJ 1015] Jury Compromise

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 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.

题解

? 第一反应是f[i][j][k][x]表示最后一个是第i个，上一个是第j个，已选k个，sum（D(J) - P(J)）== x 的最大的sum（D(J) - P(J)）。O（200*200*20*40*20）时间复杂度直接爆炸，写到一半就收手了。其实仔细再想想应该能想到正解，但是有点浮躁就直接去查了

? 正解是f[i][j]表示已选i人，sum（D(J) - P(J)）== j的最大的sum（D(J) - P(J)）。但是这样的问题是不知道已经选过哪些人，处理方法是维护一个add_num[i][j]记录已选i人，sum（D(J) - P(J)）== j的最后新加的人的编号，每次都沿着数组往回走，最后一个人是x = add_num[i][j],则上一个人是add[i-1][j-(d[x] - p[x])]。O(20*40*20*200*20)

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define inf 1000000009
using namespace std;
int n,m;
struct Node{
    int d,p,pos;
}node[205];
int f[25][1005];
int add_num[25][1005];//the last number to get to this status
int sum1,sum2;
int ans_q[25];
int main(){
    int t = 0;
    while(~scanf("%d%d",&n,&m),n+m){
        for (int i = 1;i <= n;++i){
            scanf("%d%d",&node[i].d,&node[i].p);
            node[i].pos = i;
        }
        memset(f,-1,sizeof(f));
        memset(add_num,-1,sizeof(add_num));
        f[0][20*m] = 0;
        for (int i = 0;i < m;++i){
            for (int j = 0;j <= 40*m;++j){
                if (f[i][j] == -1) continue;//not exist
                for (int k = 1;k <= n;++k){
                    int m_num = i,sum_num = j;
                    bool k_exist = false;
                    while(m_num != 0){
                        if (add_num[m_num][sum_num] == k){
                            k_exist = true;
                            break;
                        }else{
                            int tmp = add_num[m_num][sum_num];
                            if (tmp == -1) break;
                            sum_num -= (node[tmp].d - node[tmp].p);
                            m_num--;
                        }
                    }
                    if (k_exist) continue;
                    int new_j = j + node[k].d - node[k].p;
                    if (f[i+1][new_j] == -1 || f[i][j] + node[k].d + node[k].p > f[i+1][new_j]){
                        f[i+1][new_j] = f[i][j] + node[k].d + node[k].p;
                        add_num[i+1][new_j] = k;
                    }
                }
            }
        }
        int ans_j = 0;
        while(f[m][m*20 + ans_j] == -1 && f[m][m*20 - ans_j] == -1) ans_j++;
        if (f[m][m*20 + ans_j] < f[m][m*20 - ans_j]){
            ans_j = m*20 - ans_j;
        }else ans_j = m*20 + ans_j;
        sum1 = sum2 = 0;
        for (int i = 0;i < m;++i){
            ans_q[i] = add_num[m-i][ans_j];
            ans_j -= (node[ans_q[i]].d - node[ans_q[i]].p);
            sum1 += node[ans_q[i]].d;
            sum2 += node[ans_q[i]].p;
        }
        sort(ans_q,ans_q+m);
        printf("Jury #%d
Best jury has value %d for prosecution and value %d for defence:
",++t,sum1,sum2);
        for (int i = 0;i < m;++i){
            printf(" %d",ans_q[i]);
        }
        printf("

");
    }
}