SPOJ：Collecting Candies（不错的DP）

Jonathan Irvin Gunawan is a very handsome living person. You have to admit it to live in this world.

To become more handsome, Jonathan the Handsome have to collect candies (no relation, indeed). In front of him, there are N candies with different level of sweetness. Jonathan will collect the candies one by one. Jonathan can collect any number of candies, but he must collect the candy in the increasing order of level of sweetness (no two candies will have the same level of sweetness).

Every candy has their own color, which will be represented by a single integer between 0 and 10⁹ inclusive.

If Jonathan collects the first candy, or a candy that has different color with the previous candy he take, he will get 1 point.

If Jonathan collects the candy that has the same color with the previous candy, he will get a combo. Combo-x means that he has collected x candies of the same color consecutively. In other words, if he collect a candy and get combo-(x-1) and he collect a candy with the same color again, he will get combo-(x). And then if he collects a candy with different color, the combo will vanish and back to combo- 1.

(Note : previous candy means the last candy he take)

Every time he get combo-x, he will get x points. Jonathan wants to count how many maximum total points he can get. You are a fan of Jonathan the Handsome have to help him.

Input

 

The first line consists of a single integer T, indicating the number of

testcases.

For every testcase, the first line consists of a single integer N.

The next line consists of N integers, representing the color of the candy given in the increasing level of sweetness, separated by a single space.

The first line consists of a single integer T, indicating the number of testcases.

For every testcase, the first line consists of a single integer N (1 ≤ N ≤ 1000).

The next line consists of N integers, representing the color of the candy given in the increasing level of sweetness, separated by a single space.

 

Output

For every case, output a single integer consist of the maximum total points Jonathan can get.

Example

Input:
2
4 
1 1 2 1 
4 
1 2 3 1

Output:
6
4

Explanation

In the first sample, Jonathan chooses not to take the candy with color 2, as he will lose the combo. In the second sample, he will get a maximum total points if he take all of the candies.

题意：N个数，取其子数列，使得总得分最高。得分定义如下：

       对于某一个x

       若前面有连续的c个x，则得分为c + 1

       e.g.我选择下面这些数。

     1 2 1 1 3 3 3 3 2 2 1 （取到的数列）

     1 1 1 2 1 2 3 4 1 2 1 （得分分值）

思路：DP，先离散化。

      状态F[i, j] ，i ——取到的最后一个数为i （i是离散化化后的数字），j ——前面有连续j个i ，F[i, j] ——这种情况下的最大得分

      对于当前的数now

           F[now, j + 1] = F[now, j] + j （1’）

           F[now, 1] = max{F[i, j]} + 1 (i != now) （2’）

 （感悟：平时的DP，即便是二维的DP，其状态都是一维的。所以想到还是有点难想到。读者有兴趣可以自己想试着想一下。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
int a[maxn],b[maxn],f[maxn][maxn],cnt,ans;
int main()
{
    int T,N,pos,i,j;
    scanf("%d",&T);
    while(T--){
        ans=0; scanf("%d",&N);
        for(i=1;i<=N;i++) scanf("%d",&a[i]),b[i]=a[i];
        sort(b+1,b+N+1);
        cnt=unique(b+1,b+N+1)-(b+1);
        memset(f,0,sizeof(f));
        for(i=1;i<=N;i++){
            pos=lower_bound(b+1,b+cnt+1,a[i])-b;    
            for(j=i;j>=2;j--) if(f[pos][j-1]) f[pos][j]=max(f[pos][j],f[pos][j-1]+j);
            f[pos][1]=ans+1;
            for(j=1;j<=i;j++) ans=max(ans,f[pos][j]);
        }
        printf("%d\n",ans);
    }
    return 0;
}