【HOJ2430】【贪心+树状数组】 Counting the algorithms

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

【分析】

比较水的一道题，根据题意可知，任何两个区间仅为相交或包含的关系。

相交无论先删除哪一个对结果都没有影响，包含当然要先删除外面的那一个，所以从后往前删不会有包含的关系。

具体用树状数组实现就可以了。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <map>

 const int MAXN =  *  + ;
 using namespace std;
 struct DATA{
        int l, r;
 }num[MAXN * ];
 int data[MAXN * ];
 int C[MAXN * ], cnt[MAXN * ];
 int n;
 bool get[MAXN * ];

 int lowbit(int x){return x&-x;}
 void add(int x, int val){
      while (x <=  * n){
            C[x] += val;
            x += lowbit(x);
      }
      return;
 }
 int sum(int x){
     int cnt = ;
     while (x > ){
           cnt += C[x];
           x -= lowbit(x);
     }
     return cnt;
 }
 void init(){
      C[] = ;
      for (int i = ; i <=  * n; i++) C[i] = cnt[i] = ;
      for (int i = ; i <=  * n; i++){
          scanf("%d", &data[i]);
          if (cnt[data[i]] == ) {num[data[i]].l = i;cnt[data[i]]++;}
          else num[data[i]].r = i;
          add(i, );
      }
      long long Ans = ;
      memset(get, , sizeof(get));
      for (int i =  * n; i >= ; i--){
          if (get[i] == ) continue;
          Ans += sum(i) - sum(num[data[i]].l);
          add(i, -);
          add(num[data[i]].l, -);
          get[i] = get[num[data[i]].l] = ;
      }
      printf("%lld\n", Ans);
 }

 int main(){
      int T = ;
      #ifdef LOCAL
      freopen("data.txt", "r", stdin);
      freopen("out.txt", "w", stdout);
      #endif
      while (scanf("%d", &n) != EOF){
            init();
      }
      return ;
 }