• AtCoder Beginner Contest 116 D



    D - Various Sushi


    Time Limit: 2 sec / Memory Limit: 1024 MB

    Score : 400400 points

    Problem Statement

    There are NN pieces of sushi. Each piece has two parameters: "kind of topping" titi and "deliciousness" didi. You are choosing KK among these NN pieces to eat. Your "satisfaction" here will be calculated as follows:

    • The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".
    • The base total deliciousness is the sum of the deliciousness of the pieces you eat.
    • The variety bonus is xxx∗x, where xx is the number of different kinds of toppings of the pieces you eat.

    You want to have as much satisfaction as possible. Find this maximum satisfaction.

    Constraints

    • 1KN1051≤K≤N≤105
    • 1tiN1≤ti≤N
    • 1di1091≤di≤109
    • All values in input are integers.

    Input

    Input is given from Standard Input in the following format:

    NN KK
    t1t1 d1d1
    t2t2 d2d2
    ..
    ..
    ..
    tNtN dNdN
    

    Output

    Print the maximum satisfaction that you can obtain.


    Sample Input 1 Copy

    Copy
    5 3
    1 9
    1 7
    2 6
    2 5
    3 1
    

    Sample Output 1 Copy

    Copy
    26
    

    If you eat Sushi 1,21,2 and 33:

    • The base total deliciousness is 9+7+6=229+7+6=22.
    • The variety bonus is 22=42∗2=4.

    Thus, your satisfaction will be 2626, which is optimal.


    Sample Input 2 Copy

    Copy
    7 4
    1 1
    2 1
    3 1
    4 6
    4 5
    4 5
    4 5
    

    Sample Output 2 Copy

    Copy
    25
    

    It is optimal to eat Sushi 1,2,31,2,3 and 44.


    Sample Input 3 Copy

    Copy
    6 5
    5 1000000000
    2 990000000
    3 980000000
    6 970000000
    6 960000000
    4 950000000
    

    Sample Output 3 Copy

    Copy
    4900000016
    

    Note that the output may not fit into a 3232-bit integer type.

     题意:

    给定N个结构体,每一个结构体有两个信息,分别是type  和 x,让你从中选出K个结构体,

    使之type的类型数的平方+sum{ xi } 最大。

    思路:

    可以贪心来做。

    首先根据x由大到小来排序,然后选入结构体,先贪心的全前K个,把每一种类型的第一个一定的加入到我们选择的范围中,(贪心思想)

    然后把某一种类型的后面几个的结构体加入到栈中,

    然后扫k+1~n的时候,如果这个结构体的类型没加入过,那么把他加入,然后类型数目+1,弹出栈顶的元素,减去他的x值贡献,然后用新结构体取尝试更新最大值。

    还主要用到了stack的先进后出的思想,巧妙的最优的替换了每一个能替换掉的是当前选择中贡献最小的。

    贪心好题,(口胡结束)。细节见代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <vector>
    #define rt return
    #define sz(a) int(a.size())
    #define all(a) a.begin(), a.end()
    #define rep(i,x,n) for(int i=x;i<n;i++)
    #define repd(i,x,n) for(int i=x;i<=n;i++)
    #define pii pair<int,int>
    #define pll pair<long long ,long long>
    #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
    #define MS0(X) memset((X), 0, sizeof((X)))
    #define MSC0(X) memset((X), '', sizeof((X)))
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define eps 1e-6
    #define gg(x) getInt(&x)
    #define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
    using namespace std;
    typedef long long ll;
    ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
    ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
    inline void getInt(int* p);
    const int maxn=1000010;
    const int inf=0x3f3f3f3f;
    /*** TEMPLATE CODE * * STARTS HERE ***/
    int n,k;
    struct info
    {
        int x,v;
    }a[maxn];
    bool cmp(info one,info two)
    {
        return one.v>two.v;
    }
    int vis[maxn];
    stack<int> s;
    int main()
    {
        scanf("%d %d",&n,&k);
        repd(i,1,n)
        {
            scanf("%d %d",&a[i].x,&a[i].v);
        }
        sort(a+1,a+1+n,cmp);
        ll cnt=0;
        ll tp=0;
        ll res=0;
        ll ans=0ll;
        repd(i,1,n)
        {
            if(cnt<k)
            {
                    if(vis[a[i].x]==0)
                    {
                        vis[a[i].x]=1;
                        tp++;
                    }else
                    {
                        s.push(a[i].v);
                    }
                    res+=a[i].v;
                    cnt++;
                    ans=max(ans,res+1ll*tp*tp);
            }
            else{
                if(s.empty())
                    break;
                if(vis[a[i].x])
                    continue;
                vis[a[i].x]=1;
                tp++;
                res-=s.top();
                res+=a[i].v;
                s.pop();
                ans=max(ans,res+tp*tp);
            }
        }
        printf("%lld
    ",ans);
        return 0;
    }
    
    inline void getInt(int* p) {
        char ch;
        do {
            ch = getchar();
        } while (ch == ' ' || ch == '
    ');
        if (ch == '-') {
            *p = -(getchar() - '0');
            while ((ch = getchar()) >= '0' && ch <= '9') {
                *p = *p * 10 - ch + '0';
            }
        }
        else {
            *p = ch - '0';
            while ((ch = getchar()) >= '0' && ch <= '9') {
                *p = *p * 10 + ch - '0';
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/qieqiemin/p/10338363.html
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