codeforces-455A

最新推荐文章于 2021-10-28 14:01:16 发布

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最新推荐文章于 2021-10-28 14:01:16 发布

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原文链接：http://www.cnblogs.com/cryingrain/p/8366780.html

本文介绍了一道来自CodeForces的A. Boredom题目，通过动态规划的方法解决了一个关于序列操作的游戏问题，旨在获取最大得分。文章详细解析了状态定义、转移方程，并提供了完整的C++代码实现。

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题目连接：http://codeforces.com/contest/455/problem/A

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.

题目大意：有n个数，选择一个（设为ak）删除可以得ak分，但是数列中所有等于ak+1或ak-1的数必须删去，求最多可以得多少分。

很典型的dp题目，我比较菜所以状态设置为比较好理解的 dp[i][0]代表不选i的最大值，dp[i][1]代表选i得到的最大值。

转移方程就很好写了，边界也很清晰，详见代码

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;

LL dp[100005][2];
LL c[100005]={0};
int main()
{
    int n;
    int in;
    int z=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&in);
        c[in]++;
        z=max(z,in);
    }
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=z;i++)
    {
        dp[i][1]=dp[i-1][0]+c[i]*LL(i);
        dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
    }
    cout<<max(dp[z][0],dp[z][1]);
}