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

[kuangbin Basic dp] [POJ 1015] Jury Compromise

제목.

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.

문제풀이

첫 번째 반응은 f[i][j][k][x]가 마지막은 i번째, 마지막은 j번째, 이미 k개를 선택했고sum(D(J)-P(J))==x의 가장 큰sum(D(J)-P(J)))))이다.O(200*200*20*40*20)시간의 복잡도는 직접 폭발하여 절반까지 쓰고 손을 뗐다.사실 곰곰이 생각해 보면 정해가 떠오르겠지만 조금 경솔해서 바로 찾아봤어요.
정해는 f[i][j]가 i인을 선택했음을 의미하며,sum(D(J)-P(J))==j의 최대sum(D(J)-P(J))))이다.그러나 이런 문제는 이미 어떤 사람을 뽑았는지 모르겠고, 처리 방법은 하나의dd 를 유지하는 것이다.num[i][j]는 선택한 i인을 기록하고sum(D(J)-P(J))==j의 마지막으로 새로 추가된 사람의 번호를 기록합니다. 매번 그룹을 따라 돌아갑니다. 마지막 사람은 x=add 입니다.num[i][j], 이전 사람은dd[i-1][j-(d[x]-p[x])]입니다.O(20*40*20*200*20)

코드

#include 
#include 
#include 
#include 
#include 
#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("

");
    }
}