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。
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <vector>
6 #include <utility>
7 #include <iomanip>
8 #include <string>
9 #include <cmath>
10 #include <map>
11
12 const int MAXN = 100000 * 2 + 10;
13 using namespace std;
14 struct DATA{
15 int l, r;
16 }num[MAXN * 2];
17 int data[MAXN * 2];
18 int C[MAXN * 2], cnt[MAXN * 2];
19 int n;
20 bool get[MAXN * 2];
21
22 int lowbit(int x){return x&-x;}
23 void add(int x, int val){
24 while (x <= 2 * n){
25 C[x] += val;
26 x += lowbit(x);
27 }
28 return;
29 }
30 int sum(int x){
31 int cnt = 0;
32 while (x > 0){
33 cnt += C[x];
34 x -= lowbit(x);
35 }
36 return cnt;
37 }
38 void init(){
39 C[0] = 0;
40 for (int i = 1; i <= 2 * n; i++) C[i] = cnt[i] = 0;
41 for (int i = 1; i <= 2 * n; i++){
42 scanf("%d", &data[i]);
43 if (cnt[data[i]] == 0) {num[data[i]].l = i;cnt[data[i]]++;}
44 else num[data[i]].r = i;
45 add(i, 1);
46 }
47 long long Ans = 0;
48 memset(get, 0, sizeof(get));
49 for (int i = 2 * n; i >= 1; i--){
50 if (get[i] == 1) continue;
51 Ans += sum(i) - sum(num[data[i]].l);
52 add(i, -1);
53 add(num[data[i]].l, -1);
54 get[i] = get[num[data[i]].l] = 0;
55 }
56 printf("%lld\n", Ans);
57 }
58
59 int main(){
60 int T = 0;
61 #ifdef LOCAL
62 freopen("data.txt", "r", stdin);
63 freopen("out.txt", "w", stdout);
64 #endif
65 while (scanf("%d", &n) != EOF){
66 init();
67 }
68 return 0;
69 }
70