Codeforces Round #260 (Div. 2) C （DP）

最新推荐文章于 2025-06-01 18:11:54 发布

原创最新推荐文章于 2025-06-01 18:11:54 发布 · 858 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm #acm #codeforces #sizeof #iostream

DP 专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个游戏中的得分策略问题，玩家需要从一系列整数中选择元素并删除相邻的元素以获得最大得分，通过动态规划算法解决该问题。

题目：

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

input

3
1 2 3

output

input

9
1 2 1 3 2 2 2 2 3

output

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.

题意：

题意还是比较明显，就是在一串数字中选取一个数ak，然后把ak-1，ak+1相应的删除掉，可以获得ak的points，然后求最大能得多少points。

比较明显的DP。状态转移方程：

dp[i][0] = max(dp[i-1][0],dp[i-1][1]) ;

dp[i][1] = dp[i-1][0] + (long long)w[i]*i ;

w[i]数组用作记录每个值的个数。

dp[i][0]表示当前这个不取，dp[i][1]表示取。然后取的时候，前面位不取并且加上值就好了。

PS：这题就有个坑就是w[i]*i的时候可能会爆int。所以要加long long。这题的cha点貌似也在这里

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
int a[1111111] ;
int w[1111111] ;
long long max(long long a,long long b)
{
	return a>b?a:b;
}
long long dp[1111111][2] ;
int main()
{
	int i ;
	int n ;
	scanf("%d",&n) ;
	memset(w,0,sizeof(w)) ;
	for(i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		w[a[i]]++ ;
	}
	//cout<<a[1]<<endl;
	if(n==1)
	{
		cout<<a[1]<<endl;
		return 0 ;
	}
	memset(dp,0,sizeof(dp)) ;
	dp[1][0] = 0 ;
	dp[1][1] = w[1] ;

	for(i=2;i<=100011;i++)
	{
		dp[i][0] = max(dp[i-1][0],dp[i-1][1]) ;
		dp[i][1] = dp[i-1][0] + (long long)w[i]*i ;
	}
	cout<<max(dp[100011][0],dp[100011][1])<<endl;
	return 0 ;
}