Codeforces #260 (Div. 2)　C. Boredom(动态规划)

最新推荐文章于 2022-12-16 16:19:21 发布

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最新推荐文章于 2022-12-16 16:19:21 发布

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分类专栏：动态规划

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本文介绍了一道关于游戏策略的动态规划题目，玩家通过删除序列中的元素获取分数，目标是获得尽可能多的分数。文章提供了正确的实现代码，并解释了算法背后的逻辑。

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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.

一道DP题，自己写的总是跪在12个样例

12个样例的测试数据是100,000,导致我差错也差不了

我的想法是用a[i]存储i出现的次数与i的乘积

因为每相邻两项总要取其中一项,所以有a[i] += max(a[i-2], a[i-3])

比如:1 2 3 4 5 6

可以选取:

1 3 5

1 3 6

1 4 6

2 4 6

循环遍历如果有a[i]!=0,则取a[i] += max(a[i-2], a[i-3]);

结果错了,想不通怎么回事

先附上正确的答案吧:

#include <cstdio>
#include <iostream>
#define MAXN 100100
#define LL long long
using namespace std;

LL n, k, a[MAXN];

int main(void) {
    cin >> n;
    for(LL i=0; i<n; ++i) {
        cin >> k;
        a[k]++;
    }
    //a[i]表示前i个回合得分最大值
    for(LL i=2; i<100001; ++i) {
        a[i] = max(a[i-2]+i*a[i], a[i-1]);//要或不要第i-1项
        //cout << "a[" << i << "] = " << a[i] << endl;
    }
    cout << a[100000] << endl;
    return 0;
}