OpenJudge 2979 陪审团的人选 / Poj 1015 Jury Compromise

1.链接地址：

http://bailian.openjudge.cn/practice/2979

http://poj.org/problem?id=1015

2.题目：

总Time Limit:

1000ms

Memory Limit:

65536kB

Description

在遥远的国家佛罗布尼亚，嫌犯是否有罪，须由陪审团决定。陪审团是由法官从公众中挑选的。先随机挑选n个人作为陪审团的候选人，然后再从这n个人中选m人组成陪审团。选m人的办法是：

控 方和辩方会根据对候选人的喜欢程度，给所有候选人打分，分值从0到20。为了公平起见，法官选出陪审团的原则是：选出的m个人，必须满足辩方总分和控方总 分的差的绝对值最小。如果有多种选择方案的辩方总分和控方总分的之差的绝对值相同，那么选辩控双方总分之和最大的方案即可。

Input

输入包含多组数据。每组数据的第一行是两个整数n和m，n是候选人数目，m是陪审团人数。注意，1<=n<=200, 1<=m<=20 而且 m<=n。接下来的n行，每行表示一个候选人的信息，它包含2个整数，先后是控方和辩方对该候选人的打分。候选人按出现的先后从1开始编号。两组有 效数据之间以空行分隔。最后一组数据n=m=0

Output

对每组数据，先输出一行，表示答案所属的组号,如 'Jury #1', 'Jury #2', 等。接下来的一行要象例子那样输出陪审团的控方总分和辩方总分。再下来一行要以升序输出陪审团里每个成员的编号，两个成员编号之间用空格分隔。每组输出数据须以一个空行结束。

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 

Source

Southwestern European Regional Contest 1996, POJ 1015, 程序设计实习2007

3.思路：

这题自己没想到思路，参照了《程序设计导引及在线实践》的思路

动态规划题目，使用深度搜索的话会超时

4.代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

const int maxDiff = 400;

int cmp(const void* a,const void *b)
{
    int int_a = *((int *)a);
    int int_b = *((int *)b);

    return int_a - int_b;
}

int main()
{
    //freopen("C://input.txt","r",stdin);

    int i,j,k;
    int temp;
    int temp_path,temp_diff;

    int n,m;
    cin >> n >> m;

    int count = 1;
    while(n != 0 || m != 0)
    {
        int *arr_sum = new int[n];
        int *arr_diff = new int[n];

        bool *arr_used = new bool[n];
        memset(arr_used,0,sizeof(bool) * n);

        int int_a,int_b;
        for(i = 0; i < n; ++i)
        {
            cin >> int_a >> int_b;
            arr_sum[i] = int_a + int_b;
            arr_diff[i] = int_a - int_b;
        }

        int **dp = new int*[m];
        for(i = 0; i < m; ++i)
        {
            dp[i] = new int[maxDiff * 2 + 1];
            for(j = 0; j <= maxDiff * 2; ++j) dp[i][j] = -1;
        }

        int **path = new int*[m];
        for(i = 0; i < m; ++i) path[i] = new int[maxDiff * 2 + 1];

        for(i = 0; i < n; ++i)
        {
            temp = arr_diff[i] + maxDiff;
            if(dp[0][temp] < arr_sum[i])
            {
                dp[0][temp] = arr_sum[i];
                path[0][temp] = i;
            }

            //cout << "dp[0][" << temp << "]="  << dp[0][temp] << "path[0][" << temp << "]=" << path[0][temp] << endl;
        }


        for(i = 1; i < m; ++i)
        {
            for(j = 0; j <= maxDiff * 2; ++j)
            {
                if(dp[i - 1][j] != -1)
                {

                    temp_diff = j;
                    for(k = i - 1; k >= 0; --k)
                    {
                        temp_path = path[k][temp_diff];
                        arr_used[temp_path] = true;
                        temp_diff = temp_diff - arr_diff[temp_path];
                    }

                    for(k = 0; k < n; ++k)
                    {
                        if(!arr_used[k])
                        {
                            temp = j + arr_diff[k];
                            if(dp[i][temp] == -1 || (dp[i][temp] < dp[i - 1][j] + arr_sum[k]))
                            {
                                dp[i][temp] = dp[i - 1][j] + arr_sum[k];
                                path[i][temp] = k;
                            }
                        }
                    }


                    temp_diff = j;
                    for(k = i - 1; k >= 0; --k)
                    {
                        temp_path = path[k][temp_diff];
                        arr_used[temp_path] = false;
                        temp_diff = temp_diff - arr_diff[temp_path];
                    }
                }
            }

        }

        int res_idx;
        for(i = 0; i <= maxDiff; ++i)
        {
            if(dp[m - 1][maxDiff - i] != -1 || dp[m - 1][maxDiff + i] != -1)
            {
                if(dp[m - 1][maxDiff - i] == -1) res_idx = i;
                else if(dp[m - 1][maxDiff + i] == -1) res_idx = - i;
                else
                {
                    res_idx = (dp[m - 1][maxDiff + i] > dp[m - 1][maxDiff - i] ? i : (-i));
                }
                break;
            }
        }

        cout << "Jury #" << count++ << endl;
        cout << "Best jury has value " << (dp[m - 1][res_idx + maxDiff] + res_idx) / 2<< " for prosecution and value " << (dp[m - 1][res_idx + maxDiff] - res_idx) / 2<< " for defence:" << endl;

        temp_diff = res_idx + maxDiff;
        int *arr_res = new int[m];
        for(i = m - 1; i >= 0; --i)
        {
            temp_path = path[i][temp_diff];
            arr_res[i] = temp_path + 1;
            temp_diff = temp_diff - arr_diff[temp_path];
        }
        qsort(arr_res,m,sizeof(int),cmp);

        for(i = 0; i < m; ++i) cout << " " << arr_res[i];
        cout << endl;

        delete [] arr_res;

        cout << endl;

        delete [] arr_sum;
        delete [] arr_diff;
        
        for(i = 0; i < m; ++i) delete [] dp[i];
        delete [] dp;



        for(i = 0; i < m; ++i) delete [] path[i];
        delete [] path;

        cin >> n >> m;
    }

    return 0;
}