poj1015 Jury Compromise【背包】

Jury Compromise

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:32355		Accepted:8722		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

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

题意：

从n个候选人中选出m个，每个人有两个分数，分别是辩方和控方打出的。现在希望选出的这m个人，他们的辩方分数和与控方分数和之差的绝对值最小，当有多种情况时选择两个分数和最大的一种。还要输出方案。

思路：

感觉略难。

我们可以把n个候选人当做是n个物品，每个人的人数作为一维体积，装满容积为m的背包。每个候选人辩、控得分差作为体积之一，辩、控双方的得分和作为价值。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。并且，我们还规定，如果没法选j 个人，使其辩控差为k，那么f(j, k)的值就为-1，也称方案f(j, k)不可行。本题是要求选出m 个人，那么，如果对k 的所有可能的取值，求出了所有的f(m, k) (-20×m≤ k ≤ 20×m)，那么陪审团方案自然就很容易找到了。可行方案f(j-1, x)能演化成方案f(j, k)的必要条件是：存在某个候选人i，i 在方案f(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f(j-1, x)中，选出 f(j-1, x) + S(i) 的值最大的那个，那么方案f(j-1, x)再加上候选人i，就演变成了方案 f(j, k)。这中间需要将一个方案都选了哪些人都记录下来。不妨将方案f(j, k)中最后选的那个候选人的编号，记在二维数组的元素path[j][k]中。那么方案f(j, k)的倒数第二个人选的编号，就是path[j-1][k-V[path[j][k]]。假定最后算出了解方案的辩控差是k，那么从path[m][k]出发，就能顺藤摸瓜一步步求出所有被选中的候选人。

 //#include <bits/stdc++.h>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<stdio.h>
 #include<cstring>
 #include<map>

 #define inf 0x3f3f3f3f
 using namespace std;
 typedef long long LL;

 int n, m;
 const int maxn = ;
 const int maxm = ;
 int p[], d[], ans[];
 int dp[][], path[][];

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

         for(int i = ; i <= n; i++){
             scanf("%d%d", &p[i], &d[i]);
         }

         memset(dp, -, sizeof(dp));
         memset(path, , sizeof(path));
         dp[][ * m] = ;
         for(int j = ; j < m; j++){//j表示选出的人的数目
             for(int k = ; k <= m * ; k++){
                 if(dp[j][k] >= ){//方案（j，k）可行
                     for(int i = ; i <= n; i++){//找i是否出现过并且是否值得更新
                         int t1, t2;
                         if(dp[j][k] + p[i] + d[i] > dp[j + ][k + p[i] - d[i]]){
                             t1 = j; t2 = k;
                             while(path[t1][t2] != i && t1 > ){
                                 t2 -= p[path[t1][t2]] - d[path[t1][t2]];
                                 t1--;
                             }
                             if(t1 == ){
                                 dp[j + ][k + p[i] - d[i]] = dp[j][k] + p[i] + d[i];
                                 path[j + ][k + p[i] - d[i]] = i;
                             }
                         }
                     }
                 }
             }
         }

         int x = m * , y = ;
         while(dp[m][x + y] <  && dp[m][x - y] < )y++;
         int k;
         if(dp[m][x + y] > dp[m][x - y]){
             k = x + y;
         }
         else{
             k = x - y;
         }

         printf("Jury #%d\n",cas++);
         printf("Best jury has value %d for prosecution and value %d for defence:\n",(k-m*+dp[m][k])/,(dp[m][k]-k+m*)/);
         for(int i=;i<=m;i++)
         {
             ans[i]=path[m-i+][k];
             k-=p[ans[i]]-d[ans[i]];
         }
         sort(ans + , ans + m + );
         for(int i=;i<=m;i++)
             printf(" %d",ans[i]);
         printf("\n\n");
     }
     return ;
 }