?

Counting the algorithms

Submitted : 701, Accepted : 273

As most of the ACMers, wy's next target is algorithms, too. wy is clever, so he can learn most of the algorithms quickly. After a short time, he has learned a lot. One day, mostleg asked him that how many he had learned. That was really a hard problem, so wy wanted to change to count other things to distract mostleg's attention. The following problem will tell you what wy counted.

Given 2N integers in a line, in which each integer in the range from 1 to N will appear exactly twice. You job is to choose one integer each time and erase the two of its appearances and get a mark calculated by the differece of there position. For example, if the first 3 is in position 86 and the second 3 is in position 88, you can get 2 marks if you choose to erase 3 at this time. You should notice that after one turn of erasing, integers' positions may change, that is, vacant positions (without integer) in front of non-vacant positions is not allowed.

Input

There are multiply test cases. Each test case contains two lines.

The first line: one integer N(1 <= N <= 100000).

The second line: 2N integers. You can assume that each integer in [1,N] will appear just twice.

Output

One line for each test case, the maximum mark you can get.

Sample Input

?

3
1 2 3 1 2 3
3
1 2 3 3 2 1

Sample Output

?

6
9

Hint

We can explain the second sample as this. First, erase 1, you get 6-1=5 marks. Then erase 2, you get 4-1=3 marks. You may notice that in the beginning, the two 2s are at positions 2 and 5, but at this time, they are at positions 1 and 4. At last erase 3, you get 2-1=1marks. Therefore, in total you get 5+3+1=9 and that is the best strategy.

?

?

————————————————————分割线——————————————————

题目大意：

给定2n个整数，范围为1到n，每个数出现两次，每个数都有一个下标，每次删除两个相同的数，能获得下标差的值，求这个值得最大

?

思路：

对于相间的情况，从外围往内删跟从内往外删是一样的，而对于 包含的情况，从外往内删是最优的。因此，选择从外往内删。

具体操作：

扫描一次，记录每个数第二次出现的下标，将数的下标插人树状数组。然后从左往右或者从右往左扫描，那么值为两个数的距离差，之后删除其中一个数，如果是从左往右扫描的，则删除第二次出现的下标，如果是从右往左扫描的，则删除第一次出现的下标

?

?

?

#include
  
   
#include
   
     #include
    
      #define ll long long #define M 200000+10 using namespace std; int a[M],c[M],f[M]; bool judge[M]; int n; void update(int x,int v) { for(int i=x;i<=2*n;i+=i&-i){ c[i]+=v; } } int getsum(int x) { int sum=0; for(int i=x;i>0;i-=i&-i){ sum+=c[i]; } return sum; } int main() { while(scanf("%d",&n)!=EOF){ memset(judge,false,sizeof(judge)); memset(c,0,sizeof(c)); memset(f,0,sizeof(f)); for(int i=1;i<=2*n;++i){ scanf("%d",&a[i]); update(i,1); if(!judge[a[i]]) judge[a[i]]=true; else f[a[i]]=i; } int sum=0; for(int i=1;i<=2*n;++i){ sum+=getsum(f[a[i]])-getsum(i); update(f[a[i]],-1); } printf("%d\n",sum); } return 0; } 
    
   
  

?