Boredom - CodeForces #260 (Div. 2) C dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38459061

本文探讨了在特定游戏环境中，通过策略性选择和操作，最大化得分的方法。具体介绍了游戏规则、得分机制以及如何利用排序、动态规划等算法优化游戏体验，实现高分目标。

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Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

Sample test(s)

input
2
1 2

output
2

input
3
1 2 3

output
4

input
9
1 2 1 3 2 2 2 2 3

output
10

Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

思路：先按数大小从小到大排序，dp[i][0]表示第i个数不取时的最大得分，dp[i][1]表示第i个数取时的最大得分。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{ ll val;
  ll num;
}box[100010];
ll dp[100010][3],val[100010];
int main()
{ int T,t,n,m,i,j,k,num=0;
  scanf("%d",&n);
  for(i=1;i<=n;i++)
  { scanf("%d",&k);
    val[k]++;
  }
  for(i=1;i<=100000;i++)
   if(val[i]>0)
   { box[++num].val=i;
     box[num].num=val[i];
   }
  dp[1][1]=box[1].val*box[1].num;
  dp[1][0]=0;
  for(i=2;i<=num;i++)
  { if(box[i].val-box[i-1].val==1)
    { dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
      dp[i][1]=dp[i-1][0]+box[i].val*box[i].num;
    }
    else
    { dp[i][1]=max(dp[i-1][0],dp[i-1][1])+box[i].val*box[i].num;
      dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
    }
  }
  printf("%I64d\n",max(dp[num][0],dp[num][1]));
}