CF1419D2

题面:https://codeforces.com/problemset/problem/1419/D2

我宣布一件事：我是傻逼！

大致题意：给定一个数列，若有一个数比左右相邻的数都小，则其对答案的贡献为1，允许随意重新排列，求最终答案的最大值

题解：易知，若最终答案为n，则存在至少n+1个值严格大于对答案有贡献的数(我们在这里取最小的n个值)，故直接对数列进行排序即可

但是直接贪心排序后答案可能为错,应该找到最小的值比前一个MIN大且比后一个MIN大

我硬写了一个，ac了(大致思路是找出最可能出最优解的序列，再统计答案)

(代码比较丑，就当警示碑了bushi)

#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<vector>
#include<string>
#include<map>
#include<cstdio>
#include<list>
#include<stack>
#include<unordered_set>
using namespace std;
using ll = int64_t;
ll T, n, m, k;
ll tmp, a[300005];
int main()
{
    T = 1;
    //cin >> T;
    while(T--)
    {
        cin >> n;
        for (ll i = 1; i <= n;i++)
        {
            cin >> a[i];
        }
        sort(a + 1, a + 1 + n);
        m = (n - 1) / 2;
        k = n - m;
        vector<ll> MIN, MAX, ans;
        for (ll i = 1; i <= n;i++)
        {
            if(i<=m)
            {
                MIN.push_back(a[i]);
            }else
            {
                MAX.push_back(a[i]);
            }
        }
        int flag = 1;
        ll tot = 0, tot2 = 0, tot3 = 0;
        vector<ll> no;
        for (ll i = 1; i <= n;i++)
        {
            if(flag==1&&tot<MAX.size())
            {
                while(tot<MAX.size()&&((tot3<MIN.size()&&MAX[tot]<=MIN[tot3])||(ans.size()>0&&(MAX[tot]<=ans[ans.size()-1]))))
                {
                    no.push_back(MAX[tot++]);
                }
                if(tot<MAX.size())
                {
                    ans.push_back(MAX[tot++]);
                }
                flag = 0;
            }else
            {
                if(tot3<MIN.size())
                {
                    ans.push_back(MIN[tot3++]);
                }else
                {
                    if(tot<MAX.size())
                    ans.push_back(MAX[tot++]);
                    else
                        ans.push_back(no[tot2++]);
                }
                flag = 1;
            }
        }
        while(tot2<no.size())
        {
            ans.push_back(no[tot2++]);
        }
        ll num = 0;
        for (ll i = 1; i < n-1;i++)
        {
            if(ans[i]<ans[i-1]&&ans[i]<ans[i+1])
            {
                num++;
            }
        }
        cout << num << "\n";
        for (ll i = 0; i < n;i++)
        {
            cout << ans[i] << " ";
        }
        cout << "\n";
    }
    return 0;
}

只要将n为偶数时，MAX的位置向后移一位，我原本的代码也ac了..

#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<vector>
#include<string>
#include<map>
#include<cstdio>
#include<list>
#include<stack>
#include<unordered_set>
using namespace std;
using ll = int64_t;
ll T, n, m, k;
ll tmp, a[300005];
int main()
{
    T = 1;
    //cin >> T;
    while(T--)
    {
        cin >> n;
        for (ll i = 1; i <= n;i++)
        {
            cin >> a[i];
        }
        sort(a + 1, a + 1 + n);
        //m = (n - 1) / 2;
        //k = n - m;
        k = n / 2 + 1;
        m = k - 1;
        vector<ll> MIN, MAX, ans;
        for (ll i = 1; i <= n;i++)
        {
            if(i<=m)
            {
                MIN.push_back(a[i]);
            }else
            {
                MAX.push_back(a[i]);
            }
        }
        int flag = 1;
        ll tot = 0;
        for (ll i = 1; i <= n;i++)
        {
            if(flag==1)
            {
                ans.push_back(MAX[tot++]);
                flag = 0;
            }else
            {
                if(tot<=MIN.size())
                {
                    ans.push_back(MIN[tot - 1]);
                }else
                {
                    ans.push_back(MAX[tot++]);
                }
                flag = 1;
            }
        }
        ll num = 0;
        for (ll i = 1; i < n-1;i++)
        {
            if(ans[i]<ans[i-1]&&ans[i]<ans[i+1])
            {
                num++;
            }
        }
        cout << num << "\n";
        for (ll i = 0; i < n;i++)
        {
            cout << ans[i] << " ";
        }
        cout << "\n";
    }
    return 0;
}