CF 426C--Sereja and Swaps

最新推荐文章于 2021-05-01 00:49:47 发布

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最新推荐文章于 2021-05-01 00:49:47 发布

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本文深入探讨了深度学习和人工智能领域的关键算法和技术，包括卷积神经网络、循环神经网络、强化学习等，同时介绍了这些技术在实际应用中的案例。

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As usual, Sereja has array a, its elements are integers: a[1], a[2], ..., a[n]. Let's introduce notation:

$\text{[math]}$

A swap operation is the following sequence of actions:

choose two indexes i, j (i ≠ j);
perform assignments tmp = a[i], a[i] = a[j], a[j] = tmp.

What maximum value of function m(a) can Sereja get if he is allowed to perform at most k swap operations?

Input

The first line contains two integers n and k (1 ≤ n ≤ 200; 1 ≤ k ≤ 10). The next line contains n integers a[1], a[2], ..., a[n] ( - 1000 ≤ a[i] ≤ 1000).

Output

In a single line print the maximum value of m(a) that Sereja can get if he is allowed to perform at most k swap operations.

Sample test(s)

input
10 2
10 -1 2 2 2 2 2 2 -1 10

output
32

input
5 10
-1 -1 -1 -1 -1

output
-1思路:dp[i][j][k]表示i到j这一段与其他进行K次操作所能达到的最大和，只需给i到j这一段建个最小堆，外面那部分建个最大堆，每次从最小堆里拿一个和最大堆的交换即可。#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define inf 0x3f3f3f3f
#define maxn 220
int a[maxn];
int sum[maxn];
int dp[maxn][maxn][12];
inline int min(int a,int b)
{
    return a>b?b:a;
}
inline int max(int a,int b)
{
    return a>b?a:b;
}
struct Min
{
    int aa;
    Min(){}
    Min(int x)
    {
        aa = x;
    }
    bool operator < (const Min & a)const
    {
        return aa > a.aa;
    }
};
struct Max
{
    int aa;
    Max(){}
    Max(int x)
    {
        aa = x;
    }
    bool operator < (const Max & a)const
    {
        return aa < a.aa;
    }
};
priority_queue <Min> qmin;
priority_queue <Max> qmax;
int main()
{
    //freopen("in.txt","r",stdin);
    int n,K;
    while(scanf("%d%d",&n,&K)==2)
    {
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        sum[0] = 0;
        for(int i = 1;i <= n;i++)
            sum[i] = sum[i-1] + a[i];
        
        for(int i = 1;i <= n;i++)
        {
            for(int j = i;j <= n;j++)
            {
                int fuck = sum[j] - sum[i-1];
                //这个就是区间的和。
                while(!qmin.empty())    qmin.pop();
                while(!qmax.empty())    qmax.pop();
                for(int k = i;k <= j;k++)
                {
                    qmin.push(Min(a[k]));
                }
                for(int k = 1;k < i;k++)
                {
                    qmax.push(Max(a[k]));
                }
                for(int k = j+1;k <= n;k++)
                {
                    qmax.push(Max(a[k]));
                }
                for(int k = 1;k <= K;k++)
                {
                    int aa = 0,bb = 0;
                    if((!qmin.empty()) && (!qmax.empty()))
                    {
                        aa = qmin.top().aa;
                        bb = qmax.top().aa;
                        qmin.pop(); qmax.pop();
                        fuck += bb - aa;
                        qmin.push(Min(bb));
                        qmax.push(Max(aa));
                    }
                    dp[i][j][k] = fuck;
                }
            }
        }
        int ans = -inf;
        for(int i = 1;i <= n;i++)
            for(int j = i;j <= n;j++)
                for(int k = 1;k <= K;k++)
                {
                    ans = max(ans,sum[j] - sum[i-1]);
                    ans = max(ans,dp[i][j][k]);
                }
        printf("%d\n",ans);
    }
    return 0;
}
             