Paint Pearls

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

因为颜色范围比较大，可以对颜色进行离散化便于处理。然后暴力，G++可以通过。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define ll __int64
#define N 50005
const int inf=0x1f1f1f1f;
int dp[N];
struct node
{
    int val,id,col;
}a[N];
vector<int>g;
int vis[N];
bool cmpval(node a,node b)
{
    return a.val<b.val;
}
bool cmpid(node a,node b)
{
    return a.id<b.id;
}
int main()
{
    int i,j,n,cnt;
    while(scanf("%d",&n)!=-1)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i].val);
        }
        cnt=1;
        for(i=2;i<=n;i++)        //颜色一样的连续石子保留一个
            if(a[i].val!=a[i-1].val)
            a[++cnt]=a[i];
        n=cnt;
        for(i=1;i<=n;i++)  //原始顺序
            a[i].id=i;
        sort(a+1,a+n+1,cmpval);
        a[1].col=0;
        cnt=0;
        for(i=2;i<=n;i++)   //为石子颜色重新编号
        {
            if(a[i].val!=a[i-1].val)
                a[i].col=++cnt;
            else
                a[i].col=cnt;
        }
        sort(a+1,a+n+1,cmpid);
        for(i=0;i<=n;i++)
            dp[i]=inf;
        dp[0]=0;
        dp[n]=n;
        for(i=0;i<n;i++)
        {
            cnt=0;
            for(j=i+1;j<=n;j++)
            {
                if(!vis[a[j].col])
                {
                    cnt++;
                    vis[a[j].col]=1;
                    g.push_back(a[j].col);
                }
                if(dp[i]+cnt*cnt>=dp[n])  //优化
                    break;
                dp[j]=min(dp[j],dp[i]+cnt*cnt);
            }
            for(j=0;j<cnt;j++)   //优化
                vis[g[j]]=0;
            g.clear();
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}