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), ' ', 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'; } } }