HOJ——T 2430 Counting the algorithms

http://acm.hit.edu.cn/hoj/problem/view?id=2430

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

Submitted : 804, Accepted : 318

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：给你长度为2*n的序列，保证1~n中每个数会出现两次，求出相同数坐标差的和的最大值、、每次得到一个坐标差都会讲两个数从序列中删除从而改变编号

贪心+树状数组

考虑两种情况 ①当两组1~n不包含时，什么顺序删数都是等价的； ②包含时，从右向左删是最优的，可以保证差最大。 用树状数组维护坐标

#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N(200000+5);
int n,ans,x[N<<1],last[N],tr[N];

#define lowbit(x) (x&((~x)+1))
inline void Update(int i,int x)
{
    for(;i<=N;i+=lowbit(i)) tr[i]+=x;
}
inline int Query(int x)
{
    int ret=0;
    for(;x;x-=lowbit(x)) ret+=tr[x];
    return ret;
}

int main()
{
    for(;~scanf("%d",&n);ans=0)
    {
        memset(tr,0,sizeof(tr));
        memset(last,0,sizeof(last));
        for(int i=1;i<=(n<<1);i++)
        {
            scanf("%d",x+i),Update(i,1);
            last[x[i]]=i;
        }
        for(int i=1;i<=(n<<1);i++)
        {
            ans+=Query(last[x[i]])-Query(i);
            Update(last[x[i]],-1);
        }
        printf("%d
",ans);
    }
    return 0;
}