hdu5009 Paint Pearls[指针优化dp]

Paint Pearls

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3239    Accepted Submission(s): 1052

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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

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/*
f[i]表示涂完前適个所化的最小代价
f[i]=min(f[j]+num(j+1,i)^2);{0<=j<i}
双向链表优化+剪枝后时间复杂度：O(n√n) 
*/
#include<map>
#include<cstdio>
#include<iostream>
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=1e5+5;
int n,a[N];
int pre[N],nxt[N];
map<int,int>mp;
int f[N];
int main(){
    while(~scanf("%d",&n)){
        mp.clear();fill(f,f+n+1,2e9);f[0]=0;
        for(int i=1;i<=n;i++) a[i]=read();
        for(int i=0;i<=n;i++) pre[i]=i-1,nxt[i]=i+1;
        //双向链表删点好神奇 
        for(int i=1,tmp;i<=n;i++){
            if(!mp.count(a[i])) mp[a[i]]=i;
            else{
                tmp=mp[a[i]];
                nxt[pre[tmp]]=nxt[tmp];
                pre[nxt[tmp]]=pre[tmp];
                mp[a[i]]=i;                
            }
            for(int j=pre[i],cnt=0;~j;j=pre[j]){
                cnt++;
                f[i]=min(f[i],f[j]+cnt*cnt);
                if(cnt*cnt>n) break;
            }
        }
        printf("%d
",f[n]);
    }
    return 0;
}