Hi, I soved this problem. The program ran about 0.422 sec and used 768 kb , but there are people who solved it 0.050 – 0.110.How is it possible?
Here is my solution:
Let diff[i]=d[i]-p[i], sum[i]=d[i]+p[i], then the problem is to choose numbers p[1]..p[m] :
p[i] belongs 1..n
p[i]<>p[j] for i<>j;
(s1=diff[p[1]]+diff[p[2]]+…diff[p[m]])==min;
And when there are many numbers that minimize |s1| , then choose the one which maximizes s2=sum[p[1]]+sum[p[2]]+..sum[p[m]].
I used such DP interpretation:
It’s trivial to solve the problem for m=1.
If there is a solution for m=k, then we can build the solution for m=k+1 using procedure:
Let the solution for m=k be saved in an array A[i][j], where A[i][j] shows whether it’s possible to choose k candidates from the first i candidates that s1=j, and if it’s possible then C[i][j] shows s2.
Then we can build an arrays B[i][j], D[i][j] for k+1 (which have the same function as A[i][j], C[i][j] for k):
Code:
for i=1 to n do
for j=< minimum possible sum> to do
{
B[i][j]=B[i-1][j]; // do not include candidate number i
D[i][j]=D[i-1][j];
if(A[i][j-diff[i]])
if( (not B[i][j]) or (D[i][j]=0 for which (A[n][j] or A[n][-j]) == true is the minimum value for |s1|.
To get chosen candidates, on each stage save whether a candidate was included or not.
Then just make a recursion descent.
I very accuratly implemented the algorithm ( using binary array to save whether a candidate was included or not), but the program is very slow.
Is it due to algorithm or implementation?

整体最优：
从M（给定）个候选人中选出N（给定）个，
得分差为X（给定）的，得分和最高的组合

局部最优：
从m（m<=M）个候选人中选出n（n<=N）个， 得分差为X（给定）的，得分和最高的组合 考虑到问题的规模—— 选出的陪审团不超过20人，每人得分的范围是0到＋20 陪审团的得分差不超出【－400，＋400】 最多有801种可能 解这道题： 先用动态规划求出给定的得分差时，最高的得分和， 然后枚举出最小的得分差。 三维数组： Map[i][j][k] 第一维：参选人员 索引含义：M个候选人中的前i个 第二维：选出人数 索引含义：选出j个 第三维：得分差 索引含义：控辩双方得分差为k 数组元素：最优解决方案 1、将动态规划表中所有方案标记为“不存在” 2、植入种子, 标记map[0][0][0]存在， 方案为谁也不选，得分和为0 假设从前m（给定）人中选出n（n<=m,n<=N）人， 得分差为X（所有可能值）的最优方案（得分和最高） 都已找出 则： FOR i:=0 TO N-1 DO FOR j:=min(X) TO max(X) DO IF (map[m][i][j]+第m+1人)>map[m][i+1][j+diff[m+1]]
THEN
map[m][i+1][j+diff[m+1]]:=
map[m][i][j]+第m+1人;
//用已经存在的方案和第m+1个候选人组合新方案

注：diff[m+1]是第m+1个候选人的得分差

应选出的方案是
使map[M][N][i]存在,且i的绝对值最小的map[M][N][i]
如果map[M][N][+i] 和map[M][N][-i]都存在，
则选取两者中得分和较大的一个。

1、动态规划表的压缩存储
map[m][x][y]只在生成map[m+1][x][y]时有用
所以动态规划表中只用保留
第一维索引为m和m+1的数据

2、方案的存储
一个方案包括两个属性
1、候选人名单
2、得分和
当M=200时
候选人名单使用位集存储，
占用224bit(7个无符号长整形)

动态规划表占用的空间:2*21*801*8=269136字节(263k)