POJ1015---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

之前想的dp[i][j]表示前i个人里选出j个人的情况下最小的辩控差，但是由于绝对值的关系，这样是不满足最优子结构的

dp[i][j] 表示 选出i个人，辩控差为j的情况下，最大的辩控和
dp[i][j] = max (dp[i - 1][x] + sum[x]);
且cha[x] + x == j （cha[i]是第i个人的辩控差，sum[i]是第i个人的辩控和
用path[i][j] 记录 dp[i][j]里选中的人的编号，枚举第i个人，迭代回溯就可以判断这个人之前是否选过
辩控差可为负，所以加一个区间偏移量

/*************************************************************************
    > File Name: POJ1015.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月15日 星期日 16时31分01秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int dp[25][1010];
int path[25][1010];
int d[222], p[222];
int ans[22];
int sum[222], cha[222];

bool judge (int i, int j, int k)
{
    while (j > 0 && path[j][k] != i)
    {
        int x = path[j][k];
        --j;
        k -= cha[x];
    }
    return !j;
}

int main ()
{
    int n, m;
    int icase = 1;
    while (~scanf("%d%d", &n, &m))
    {
        if (!n && !m)
        {
            break;
        }
        memset (dp, -1, sizeof (dp));
        memset (path, 0, sizeof(path));
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d%d", &p[i], &d[i]);
            sum[i] = p[i] + d[i];
            cha[i] = p[i] - d[i];
        }
        int fix = m * 20;
        dp[0][fix] = 0;
        for (int j = 1; j <= m; ++j)
        {
            for (int k = 0; k <= 2 * fix; ++k)
            {
                if (dp[j - 1][k] >= 0)
                {
                    for (int i = 1; i <= n; ++i)
                    {
                        if (dp[j][k + cha[i]] < dp[j - 1][k] + sum[i])
                        {
                            if (!judge (i, j - 1, k))
                            {
                                continue;
                            }
                            dp[j][k + cha[i]] = dp[j - 1][k] + sum[i];
                            path[j][k + cha[i]] = i;
                        }
                    }
                }
            }
        }
        int x;
        for (int k = 0; k <= fix; ++k)
        {
            if (dp[m][fix - k] >= 0 || dp[m][fix + k] >= 0)
            {
                x = k;
                break;
            }
        }
        int div = dp[m][fix - x] > dp[m][fix + x] ? fix - x : fix + x;
        printf("Jury #%d\n", icase++);
        int x1 = (dp[m][div] + div - fix) / 2;
        int x2 = (dp[m][div] - div + fix) / 2;
        printf("Best jury has value %d for prosecution and value %d for defence:\n", x1, x2);
        int cnt = m;
        while (path[m][div] > 0)
        {
            ans[m] = path[m][div];
            div -= cha[path[m][div]];
            --m;
        }
        sort (ans + 1, ans + cnt + 1);
        for (int i = 1; i <= cnt; ++i)
        {
            printf(" %d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}