Paint Pearls

Problem Description

Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.

In each operation, he selects some continuous pearls and all these pearls will be painted to their target colors. When he paints a string which has k different target colors, Lee will cost k² points.

Now, Lee wants to cost as few as possible to get his ideal string. You should tell him the minimal cost.

Input

There are multiple test cases. Please process till EOF.

For each test case, the first line contains an integer n(1 ≤ n ≤ 5×10⁴), indicating the number of pearls. The second line contains a₁,a₂,...,a_n (1 ≤ a_i ≤ 10⁹) indicating the target color of each pearl.

Output

For each test case, output the minimal cost in a line.

Sample Input

3
1 3 3
10
3 4 2 4 4 2 4 3 2 2

Sample Output

2
7

题意：

　　给你一个数组，每个值代表一种颜色，每次选一个区间涂颜色，代价是区间内颜色种类数的平方，涂完所有数组，问你最小代价是多少

题解:

　　设定dp[i]为前i个数的最小代价,

　　那么转移就是dp[i] = min{dp[j]+cal(j+1,i)^2} cal计算区间内颜色种类数

　　明显超时;

　　当你从i-1遍历到0去寻找那个最小dp[j]+cal(j+1,i)^2时，有些电视可以跳跃的，那就是在k~i-1里面出现过的，就可以跳过，这个用双向链表实现

　　还有一个优化：当向前遍历时，不同个数的平方已经超过单独涂色的值 即 cal(j+1,i)^2>i 直接跳出

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
const int N = 1e5+10, M = 2e2+11, inf = 2e9, mod = 1e9+7;
int dp[N],a[N],pre[N],nex[N],n;
map<int,int >mp;

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        mp.clear();
        memset(dp,127,sizeof(dp));
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)nex[i]=i+1,pre[i]=i-1;
        dp[0]=0,pre[0]=-1;
        for(int i=1;i<=n;i++)
        {
            if(!mp[a[i]]) mp[a[i]]=i;
            else
            {
                int id = mp[a[i]];
                nex[pre[id]] = nex[id];
                pre[nex[id]] = pre[id];
                mp[a[id]] = i;
            }
            int num = 0;
            for(int j=pre[i];j!=-1;j=pre[j])
            {
                num++;
                dp[i] = min(dp[i],dp[j]+num*num);
                if(num*num>i) break;
            }
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}