AtCoder Beginner Contest 116 D - Various Sushi 【贪心+栈】

Problem Statement

There are $\mathrm{NN}$ pieces of sushi. Each piece has two parameters: "kind of topping" ${\mathrm{titi}}^{}$ and "deliciousness" ${\mathrm{didi}}^{}$. You are choosing $\mathrm{KK}$ among these $\mathrm{NN}$ 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 $\mathrm{x\ast xx\ast x}$, where $\mathrm{xx}$ 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

• $\mathrm{1\le K\le N\le 1051\le K\le N\le 105}$
• $\mathrm{1\le ti\le N1\le ti\le N}$
• $\mathrm{1\le di\le 1091\le di\le 109}$
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$ $\mathrm{KK}$
${\mathrm{t1t1}}^{}$ ${\mathrm{d1d1}}^{}$
${\mathrm{t2t2}}^{}$ ${\mathrm{d2d2}}^{}$
$..$
$..$
$..$
${\mathrm{tNtN}}^{}$ ${\mathrm{dNdN}}^{}$

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 $\mathrm{1,21,2}$ and $33$:

• The base total deliciousness is $\mathrm{9+7+6=229+7+6=22}$.
• The variety bonus is $\mathrm{2\ast 2=42\ast 2=4}$.

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 $\mathrm{1,2,31,2,3}$ 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：【贪心】将X从大到小排序，然后按顺序取前K个，在取前K个过程中，将已经出现的类型放入栈中。然后，开始遍历K+1----N的元素，使得不断加入没有出现的元素的类型。在此过程中通过弹栈更新最值。

AC代码：

#include<bits/stdc++.h>

using namespace std;
#define int long long
#define N 150000
struct str{
    int x,y;
}st[N];
bool cmp(str a,str b){
    return a.y>b.y;
}
map<int,int> mp;
stack<int> s;
signed  main(){
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>st[i].x>>st[i].y;
    }
    sort(st+1,st+1+n,cmp);
    int maxn=0;
    int type=0;
    int sum=0;
    for(int i=1;i<=k;i++){
        if(!mp[st[i].x]){
            mp[st[i].x]=1;
            type++;
        }else{
            s.push(st[i].y);
        }
        sum+=st[i].y;
        maxn=max(maxn,type*type+sum);    
    }
    for(int i=k+1;i<=n;i++){
        if(s.empty())
            break;
        if(mp[st[i].x])
            continue;
        mp[st[i].x]=1;
        type++;
        sum-=s.top();
        s.pop();
        sum+=st[i].y;
        maxn=max(maxn,type*type+sum);    
    }
    cout<<maxn;
    return 0;
}