AtCoder abc461_c Variety

AT_abc461_c [ABC461C] Variety

题目描述

There are $ N $ gems. The color (represented as an integer) of the $ i $ -th gem is $ C_i $ and its value is $ V_i $ .

Choose $ K $ gems from these $ N $ gems. Here, the chosen gems must have at least $ M $ distinct colors.

Find the maximum possible total value of the chosen gems. (Such a choice is always possible in the given input.)

输入格式

The input is given from Standard Input in the following format:

$ N $ $ K $ $ M $
$ C_1 $ $ V_1 $
$ C_2 $ $ V_2 $
$ \vdots $
$ C_N $ $ V_N $

输出格式

Output the maximum possible total value of the chosen gems as an integer.

输入输出样例 #1

输入 #1

5 3 2
1 30
1 40
1 50
2 10
3 20

输出 #1

110

输入输出样例 #2

输入 #2

5 3 3
1 30
1 40
1 50
2 10
3 20

输出 #2

80

输入输出样例 #3

输入 #3

5 5 1
4 1000000000
5 1000000000
4 1000000000
5 1000000000
4 1000000000

输出 #3

5000000000

说明/提示

Sample Explanation 1

In this sample, choose three gems from five gems. The chosen gems must have at least two distinct colors.

Choosing gems $ 2, 3, 5 $ gives colors $ 1, 1, 3 $ , which are two distinct colors. Their total value is $ 40 + 50 + 20 = 110 $ , and this is the maximum possible value.

Sample Explanation 2

The gems and the number to choose are the same as in Sample Input 1, but the chosen gems must have at least three distinct colors.

Choosing gems $ 3, 4, 5 $ gives colors $ 1, 2, 3 $ , which are three distinct colors. Their total value is $ 50 + 10 + 20 = 80 $ , and this is the maximum possible value.

Sample Explanation 3

Beware of overflow.

Constraints

• $ 1 \leq M \leq K \leq N \leq 2 \times 10^5 $
• $ 1 \leq C_i \leq N $
• $ 1 \leq V_i \leq 10^9 $
• There exist gems of at least $ M $ distinct colors.
• All input values are integers.

思路描述

比较容易想到的一题。

由于要求最大的价值，所以按价值从大到小排序。

因为至少要有 $ M $ 种宝石，所以先选出排完序后的、选出最大的、颜色互不相同的 $ M $ 颗宝石。最后再在没选过的宝石中选出 $ K-M $ 颗没选过的宝石就行了。

代码示例

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=200010;
int n,m,k,c,v,cnt;
ll ans;
pair<int,int> a[N];
set<int> st;
bool f[N];
int main() {
	scanf("%d %d %d",&n,&k,&m);
	for(int i=1;i<=n;i++) {
		scanf("%d %d",&c,&v);
		a[i].first=v;
		a[i].second=c;
	}
	sort(a+1,a+1+n);
	for(int i=n;i>=1;i--) {
		if(st.size()==m) break;
		if(st.find(a[i].second)==st.end()) {
			st.insert(a[i].second);
			f[i]=true;
			ans+=(ll)a[i].first;
			cnt++;
		}
	}
	for(int i=n;i>=1;i--) {
		if(f[i]) continue;
		if(cnt==k) break;
		ans+=(ll)a[i].first;
		cnt++;
	}
	printf("%lld",ans);
	return 0;
}