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 x∗xx∗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
- 1≤K≤N≤1051≤K≤N≤105
- 1≤ti≤N1≤ti≤N
- 1≤di≤1091≤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
5 3
1 9
1 7
2 6
2 5
3 1
Sample Output 1 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 2∗2=42∗2=4.
Thus, your satisfaction will be 2626, which is optimal.
Sample Input 2 Copy
7 4
1 1
2 1
3 1
4 6
4 5
4 5
4 5
Sample Output 2 Copy
25
It is optimal to eat Sushi 1,2,31,2,3 and 44.
Sample Input 3 Copy
6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000
Sample Output 3 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), '\0', 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=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
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,,n)
{
scanf("%d %d",&a[i].x,&a[i].v);
}
sort(a+,a++n,cmp);
ll cnt=;
ll tp=;
ll res=;
ll ans=0ll;
repd(i,,n)
{
if(cnt<k)
{
if(vis[a[i].x]==)
{
vis[a[i].x]=;
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]=;
tp++;
res-=s.top();
res+=a[i].v;
s.pop();
ans=max(ans,res+tp*tp);
}
}
printf("%lld\n",ans);
return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}