Computer College college Student Program Design Contest (' collector ') the ' s puzzle

problem DescriptionThere is a collector who own many valuable jewels. He has a problem on how to store them. There is M special boxes. Each box has a value. And each of the N jewels have 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 t He value of j-th box.
Note that a box can store more than one jewel.
Now the collector turns to your for helping him to compute the minimal sum of differences. InputThere is multiple test cases.
The first line has both integers N, M (1<=n, m<=100000).
The second line has N integers, indicating the n jewels ' values.
The third line has m integers, indicating the m boxes ' values.
Each value was no more than 10000. OutputPrint One integer, indicating the minimal sum of differences. Sample Input4 41 2 3 44 3 2 14 41 2 3 41 1 1 1 Sample Output06

We sort first, and then we use the second array as the basis for the first set of data to check their position in the second group.

Lower_bound
A total of two cases, either just well found, one on the left
Like what
2 3 4 5 7 find 3 return 2
Find 6 return 5 We don't know if it's on the left or the right. The difference is small, can be considered in
Find 8 to return to the last position

#include <stdio.h>//#include <bits/stdc++.h>#include <string.h>#include<iostream>#include<math.h>#include<sstream>#include<Set>#include<queue>#include<map>#include<vector>#include<algorithm>#include<limits.h>#defineMAXN (100000+10)#defineMAXM (100000)#defineINF 0X3FFFFFFF#defineINF 0x3f3f3f3f#defineLson l,m,rt<<1#defineRson m+1,r,rt<<1|1#defineLL Long Long#defineULL unsigned long Longusing namespacestd;intA[MAXN],B[MAXN];intMain () {intn,m;  while(cin>>n>>m) {intsum=0; inti,j;  for(i=1; i<=n;i++) {cin>>A[i]; }         for(i=1; i<=m;i++) {cin>>B[i]; } sort (A+1, a+n+1); Sort (b+1, b+m+1);  for(i=1; i<=n;i++)        {            intMid=inf; intAns=lower_bound (b +1, b+m+1, A[i])-b; //cout<<ans<<endl;            if(ans<=m) {Mid=min (Mid,abs (a[i]-B[ans])); }            if(ans>1) {Mid=min (Mid,abs (a[i]-b[ans-1])); } Sum+=mid; } cout<<sum<<Endl; }    return 0;}

Computer College college Student Program Design Contest (' collector ') the ' s puzzle