ATCoder 116 D (思维+贪心+栈）

Posted 2021-03-26 jrfr

D - Various Sushi

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $400400$ points

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.

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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.

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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 $44$.

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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.

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题意：

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

使之 nub 的类型数的平方+sum{val i } 最大。

思路：对于给定的结构体，按照val进行从大到小排序。

　　建立一个栈，用来存储贡献值小的数（先入栈的贡献值大）

　　预处理前k个结构体，若该结构体的nub出现过，则入栈。用sum1存储val的和，sum2存储nub种类数。总贡献就是sum1 + sum2的平方

　　定义一个整型变量maxn用来维护最大值　　

　　k+1到n，如果此结构体 i 的nub没有出现过，则取出栈顶元素x，用 i 来替换 x的信息，得出一个总贡献（不一定是最优解），用maxn更新一下。

　　最后输出maxn就ok了。注意当栈为空的时候就可以跳出循环了，因为后面的数贡献值肯定不会大于前面的）

　　

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf  0x3f3f3f3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl

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;}
//const int maxn = 110000;
using namespace std;

int n,k;
int vis[1000010];
struct node 
{
    int val,nub;
}a[1000010];
bool cmp(node x,node y)
{
    return x.val > y.val;
}
int main()
{

    //freopen("C:\ACM\input.txt","r",stdin);
    cin>>n>>k;
    for(int i = 1;i <= n; ++i) cin>>a[i].nub>>a[i].val;

    sort(a+1 ,a+1+n,cmp);
    
    stack<int>sta;

    ll sum1 = 0;
    ll sum2 = 0;
    //for(int i = 1;i <= n; ++i) cout<<a[i].nub<<" "<<a[i].val<<endl;
    for(int i = 1;i <= k; ++i)
    {
        if(vis[a[i].nub] == 1) sta.push(i);
        else 
        {
            vis[a[i].nub] = 1;
            sum2++;
        }
        sum1 += a[i].val;
    }
    //cout<<sum1 + sum2 * sum2<<endl;
    ll maxn = sum1 + sum2 * sum2;
    for(int i = k+1;i <= n; ++i)
    {
        
        if(sta.size() == 0) break;
        if(vis[a[i].nub] == 1) continue;
        int x = sta.top();
        sta.pop();
        sum1 = sum1 - a[x].val + a[i].val;
        sum2++;
        vis[a[i].nub] = 1;
        maxn = max(maxn,sum1 + sum2 * sum2);
    }
    cout<<maxn<<endl;

}