spoj ADAFRIEN - Ada and Friends（水题）

本文链接：https://blog.youkuaiyun.com/dgghjnjk/article/details/81455676

通过算法帮助Ada筛选不必要的开销，减少购买礼物的成本。输入包括庆祝次数及可删除好友数量，输出为节省的最大金额。

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ADAFRIEN - Ada and Friends

Ada the Ladybug has many friends. They always celebrate something and Ada has to buy them gifts. It is pretty costly so she have decided to unfriend some of them. What is the maximum of money she can spare?

Input

The first line of each test-case will contain two integers 1 ≤ Q K ≤ 105, the number of celebrations (Q) and the maximum number of friends Ada wants to unfriend (K).

The next Q lines will contain s (the name of friend to whom Ada will buy gift) and 1 ≤ E ≤ 109+1 (the expenses).

Names will contain at most 40 lowercase english letters.

Output

For each test-case, print the number of money Ada could spare.

Example Input

6 1
uvuvwevwevweonyetenyevweugwemubwemossas 10
ryuk 11
uvuvwevwevweonyetenyevweugwemubwemossas 5
fegla 3
tenshikanade 7
fegla 2

Example Output

Example Input

4 3
frodo 1
harrypotter 2
frodo 2
harrypotter 2

Example Output

Example Input

7 2
waynebot 7
lhic 4
petr 5
umnik 9
izrak 6
tourist 11
zlobobber 9

Example Output

Example Input

6 3
dufresne 5
gump 11
dufresne 3
mcmurphy 19
leon 10
dufresne 1

Example Output

#include<bits/stdc++.h>
#define MAXN 100005
using namespace std;
map<string,long long> mp;
map<string,long long> ::iterator it;
long long a[MAXN];
int main()
{
	long long k,q;
	string s;
	scanf("%lld%lld",&q,&k);
	for(long long i=0;i<q;i++)
	{
		long long tmp;
		cin>>s;
		scanf("%lld",&tmp);
		it=mp.find(s);
		if(it==mp.end())
			mp[s]=tmp;
		else
			mp[s]+=tmp;
	}
	long long cnt=0;
	for(it=mp.begin();it!=mp.end();it++)
		a[cnt++]=(*it).second;
	sort(a,a+cnt);
	long long sum=0;
	for(long long i=cnt-1,j=0;j<k&&i>=0;i--,j++)
		sum+=a[i];
	printf("%lld\n",sum);
	return 0;
}