【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 = 100000 * 2 + 10; 
using namespace std;
struct DATA{
       int l, r; 
}num[MAXN * 2];
int data[MAXN * 2];
int C[MAXN * 2], cnt[MAXN * 2];
int n;
bool get[MAXN * 2];

int lowbit(int x){return x&-x;}
void add(int x, int val){
     while (x <= 2 * n){
           C[x] += val;
           x += lowbit(x);
     }
     return;
}
int sum(int x){
    int cnt = 0;
    while (x > 0){
          cnt += C[x];
          x -= lowbit(x);
    }
    return cnt;
}
void init(){
     C[0] = 0;
     for (int i = 1; i <= 2 * n; i++) C[i] = cnt[i] = 0;
     for (int i = 1; i <= 2 * n; i++){
         scanf("%d", &data[i]);
         if (cnt[data[i]] == 0) {num[data[i]].l = i;cnt[data[i]]++;}
         else num[data[i]].r = i;
         add(i, 1);
     }
     long long Ans = 0;
     memset(get, 0, sizeof(get));
     for (int i = 2 * n; i >= 1; i--){
         if (get[i] == 1) continue;
         Ans += sum(i) - sum(num[data[i]].l); 
         add(i, -1);
         add(num[data[i]].l, -1);
         get[i] = get[num[data[i]].l] = 0;
     }
     printf("%lld
", Ans);
}

int main(){
     int T = 0; 
     #ifdef LOCAL
     freopen("data.txt", "r", stdin);
     freopen("out.txt", "w", stdout); 
     #endif 
     while (scanf("%d", &n) != EOF){
           init();
     }
     return 0;
}