Hoj2430 Counting the algorithms

Hoj2430 Counting the algorithms

My Tags	(Edit)

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

Submitted: 725, Accepted: 286

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 first3 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 shoshould 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

 

31 2 3 1 2 331 2 3 3 2 1

Sample Output

 

69

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 2 s 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 = 1 marks. therefore, in total you get 5 + 3 + 1 = 9 and that is the best strategy.

You can use a tree array and map Hash, to store the location where the same number appears for the second time, so it is more convenient to return when the update is needed. Then, the greedy policy is used here, that is, the loop is performed from left to right in sequence, find the same two numbers, find the difference between the two locations, and delete the two locations.

 

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          #includeusing namespace std;int a[200006],b[200006],n,vis[200006];int lowbit(int x){return x&(-x);}void update(int pos,int num){while(pos<=2*n){b[pos]+=num;pos+=lowbit(pos);}}int getsum(int pos){int num=0;while(pos>0){num+=b[pos];pos-=lowbit(pos);}return num;}int main(){int m,i,j,t,sum;while(scanf("%d",&n)!=EOF){map
          
           hash;hash.clear();for(i=1;i<=2*n;i++){vis[i]=0;scanf("%d",&a[i]);hash[a[i]]=i;b[i]=lowbit(i);}sum=0;for(i=1;i<=2*n;i++){if(vis[i]==1)continue;vis[i]=1;t=hash[a[i]];vis[t]=1;sum+=getsum(t)-getsum(i);update(i,-1);update(t,-1);//printf("%d\n",sum);}printf("%d\n",sum);}return 0;}