2015‘12杭电校赛1003 The collector’s puzzle（双指针水题）

题解：

可以说读题读不懂么…排个序然后直接双指针就可以了，水题

代码

#include <cstdio>
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <bitset>
#define MAX 100000
#define LL long long
const int INF = 999999;
using namespace std;
int cas=1,T,n,m,a[MAX+10],b[MAX+10];
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++)
scanf("%d",&b[i]);
int sum=0;
sort(a+1,a+n+1);
sort(b+1,b+m+1);
b[0]=-INF;
b[m+1] =INF;
int i=1,j=1;
while(i<=n)
{
while(a[i]>b[j])
j++;
sum+=min(abs(a[i]-b[j-1]),abs(b[j]-a[i]));
i++;
}
printf("%d\n",sum);
}
return 0;
}

题目

The collector’s puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1079 Accepted Submission(s): 244

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

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