20130712 【南华大学 ACM】 新生赛第三场 【B. A simple problem】

Problem B: A simple problem

Time Limit: 1 Sec Memory Limit:128 MB

Description

Dreamone has a lovely cat. The cat can comfort you if you are very dejected, she can also play with you if
you are very bored. So Dreamone loves her very much. Of course, you bet, she can also make you trouble if
she wants to be naughty. This time, she comes up with a simple problem for fun, just like this:
She has 2n toys. In her heart, she has used a number to stand for each toy, and for example, the 2n toys are
given like this:
A1、A2、A3、….A2n-1、A2n
She wants to divide them into two groups, where each group contains n toys. So we can use B1、
B2...Bn ,which are all from A1、 A2...A2n ,to stand for Group One ,use C1、 C2….Cn ,which are all from A1、
A2...A2n, to stand for Group Two. The cat wants to know the minimum value S for the expression below:
S=|B1-C1|+|B2-C2|+|B3-C3|+...+|Bn-Cn|
As we know Dreamone is studying MaJiang recently, so he has no time to solve the simply problem. But he
knows the 6th program contest of SWUST is on, so he turns to you for help. Can you help him?

Input

The first line of input will be a positive integer C indicating how many data sets will be included. Each of
the C data sets will contain two parts：The first part contains a number n(1<=n<=100000),and the second
parts contains 2n numbers ,which are A1、 A2、 A2n-1、 A2n (0<=Ai<=1000(1<=i<=2n)),represented 2n toys

Output

For each case, output the minimum value S for answer.

Sample Input

2
1
1 3
2
1 1 1 2

Sample Output

2
1

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解题报告：

此题不难解：

用数组存储数据，排序，相邻两项之差（升序时后项-强项；降序时前项减后项）累加。

不过我把数据当成一维数组的序号存入，这样就少了排序这步。

我的解法：

输入的数有重复时：重复次数为偶数，把该数删去； 为奇数时，只保留一个数；
数组存储数据(数组初始化0，存储数据i时，则a[i]=1，这样可按顺序存储数据)；
处理数组a的数据(保留为“1”项，删去为“0”项)，存入数组aa中（数据单调存入）；
然后相邻两项作差（后项-前项）。

代码如下：

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代码：

#include<stdio.h>
int main()
{
int N,n,i,j,x,sum,aa[1005];
scanf("%d",&N);
while(N--)
{
int a[1005]={0};
scanf("%d",&n);
n=2*n;
for(i=0;i<n;i++)
{
scanf("%d",&x);
a[x]=(a[x]+1)&1;
}
for(x=i=0;i<1000;i++)
if( a[i] )
aa[x++]=i;
for(sum=0,i=0,j=1; j<x ;i+=2,j+=2)
sum+=aa[j]-aa[i];
printf("%d\n",sum);
}
return 0;
}