hoj 2430 Counting the algorithms

Counting the algorithms

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	Source : mostleg
	Time limit : 1 sec		Memory limit : 64 M

Submitted : 484, Accepted : 186

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 3is 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.

 题意：给出2*N的序列，每个数∈[1,N]出现2次  2个数之间的间隔为得分，

    求得一个得分后会删除这两个数，问最大得分

      N比较大，应该是贪心

     从后往前删除数，用树状数组求得sum(x)个数

    对于嵌套的、相互独立的，先删除谁都没关系

    但对于包含关系的，必须先删除外围的，所以从后开始（从前开始也一样）#include <stdio.h>
#include <string.h>
#define maxn  200500
int c[maxn],be[maxn],en[maxn],a[maxn];
int n;
int lowbit( int x)
{
 return x&(-x);
}
void updata( int N,int p)
{
  
  while(N<=2*n)
  {
        c[N]+=p;N+=lowbit(N);
  }

}
int sum( int x)
{
  int s=0;
  while(x!=0)
  {
        s+=c[x];x-=lowbit(x);
  }
  return s;
}
int main( )
{
 int i;
 while(scanf("%d",&n)!=EOF)
 {
        memset(be,0,sizeof(be));
   for( i=1;i<=2*n;i++)
   {
         scanf("%d",&a[i]);
         if(be[a[i]]) en[a[i]]=i;
         else be[a[i]]=i;
         updata(i,1);
   }
   int ans=0;
   for(i=2*n;i>=1;i--)
   {
         if(be[a[i]]==0) continue;
         ans+=sum(i)-sum(be[a[i]]);
         updata(be[a[i]],-1);
         updata(i,-1);
         be[a[i]]=0;
   }
   printf("%d\n",ans);
 }
 return 0;
}
链接：http://acm.hit.edu.cn/hoj/problem/view?id=2430