计算机学院大学生程序设计竞赛（2015’12）The collector’s puzzle

The collector’s puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 944    Accepted Submission(s): 210

Problem Description

There is a collector who own many valuable jewels. He has a problem about how to store them. There are M special boxes. Each box has a value. And each of the N jewels has a value too. 
The collector wants to store every jewel in one of the boxs while minimizing the sum of difference value. 
The difference value between a jewel and a box is: |a[i] - b[j]|, a[i] indicates the value of i-th jewel, b[j] indicates the value of j-th box.
Note that a box can store more than one jewel.

Now the collector turns to you for helping him to compute the minimal sum of differences.

 

Input

There are multiple test cases.
For each case, the first line has two integers N, M (1<=N, M<=100000).
The second line has N integers, indicating the N jewels’ values.
The third line have M integers, indicating the M boxes’ values.
Each value is no more than 10000.

 

Output

Print one integer, indicating the minimal sum of differences.

 

Sample Input

   
   
   
   

    
    
    
    4 4
1 2 3 4
4 3 2 1
4 4
1 2 3 4
1 1 1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
6
   
   
   
   

 

总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int a[10005],b[10005];
int main()
{
    int N,M,t;
    while(~scanf("%d%d",&N,&M))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0; i<N; ++i)
        {
            scanf("%d",&t);
            ++a[t];
        }
        for(int i=0; i<M; ++i)
        {
            scanf("%d",&t);
            ++b[t];
        }
        long long tol=0;
        for(int i=0; i<10005; ++i)
            if(a[i]>0)
            {
                int j;
                for(j=0;; ++j)
                {
                    if(i-j>0&&b[i-j]>0)break;
                    if(i+j<10005&&b[i+j]>0)break;
                }
                tol=tol+a[i]*j;
            }
        printf("%lld\n",tol);
    }
    return 0;
}

@执念  "@☆但求“❤”安★ 下次我们做的一定会更好。。。。
为什么这次的题目是英文的。。。。QAQ...