HDOJ 5009 Paint Pearls

DP + 优化 ，因为花费是n^2的，所以num×num 大于 DP【i】的时候就可以跳出了。。。。

Paint Pearls

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1245    Accepted Submission(s): 395

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 k ² 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 ≤ 5×10 ⁴), indicating the number of pearls. The second line contains a ₁,a ₂,...,a _n (1 ≤ a _i ≤ 10 ⁹) 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

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

int n,a[50050],b[50050],last[50050];
int dp[50050];
map<int,int> mp;

int fa[50050];

int find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}

void bing(int x,int y)
{
	int X=find(x),Y=find(y);
	if(X==Y) return ;
	fa[Y]=X;
}

int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%d",a+i);
			b[i]=a[i];
		}
		sort(b+1,b+n+1);
		int cnt=unique(b,b+n+1)-b;
		for(int i=1;i<cnt;i++)
		{
			mp[b[i]]=i;	
		}
		for(int i=0;i<=n;i++)
		{
			a[i]=mp[a[i]];
			fa[i]=i;
			last[i]=-1;
		}

		for(int i=1;i<=n;i++)	
		{
			if(last[a[i]]!=-1)
			{
				bing(last[a[i]]-1,last[a[i]]);
			}
			last[a[i]]=i;
			dp[i]=i;
			int num=0;
			for(int j=i;j>0;j=find(j-1))
			{
				num++;
				if(num*num>dp[i]) break;
				int next=find(j-1);
				dp[i]=min(dp[i],dp[next]+num*num);
			}
		}

		printf("%d\n",dp[n]);
	}
	return 0;
}