| 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 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.
这道题可以用树状数组做，用map<int,int>hash,来储存相同的数第二次出现的位置，这样待会更新的时候回比较方便，然后这里用到了贪心策略，即依次从左到右进行循环，找出相同的两个数，然后求出两个位置的差，然后删除这两个位置。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using 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<int,int>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
",sum);
}
printf("%d
",sum);
}
return 0;
}