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): 986    Accepted Submission(s): 217

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 41 2 3 44 3 2 14 41 2 3 41 1 1 1

 

Sample Output

06
千万不要读错题，盒子的容量是无限的

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <queue>#include <stack>#include <set>#include <map>#include <list>#define LL long long#define INF 0x3f3f3f3fusing namespace std;int a[100100],b[100100];int main(){    int n,m;    while(cin>>n>>m)    {        for(int i=0;i<n;i++)            cin>>a[i];        for(int i=0;i<m;i++)            cin>>b[i];        sort(a,a+n);        sort(b,b+m);        int ans = 0;        int i=0, j=0;        while(i<n && j<m)        {            if(a[i] <= b[j])            {                if(j >= 1)                {                    if(a[i] - b[j-1] >= b[j] - a[i])                        ans += (b[j] - a[i++]);                    else                        ans += (a[i++] - b[j-1]);                }                else                    ans += (b[j]-a[i++]);            }            else j++;        }        while(i < n)            ans += fabs(a[i++] - b[m-1]);        cout<<ans<<endl;    }    return 0;}