Description
Winter in Yekaterinburg is the longest time of the year. And everyone spends long winter evenings in his own way. Denis is a big fan of classical music, at least twice a week he attends concerts at the Philharmonic.
This year, Denis wants to buy tickets for the whole season at once and even already has chosen a list of concerts, which he wants to attend. Now, he thinks, how to buy tickets. There is an opportunity in the Philharmonic to choose any k or more concerts and to buy a single subscription for all of them. The cost of this subscription will be equal to the total cost of all concerts in it but with a discount ppercent. There are many types of such subscriptions, with different k and p. In addition, for some concerts Denis, as a student, has a discount. This discount is valid only if he buys the ticket separately, outside the subscription.
Denis, like any other poor student, wants to attend all the concerts, spending as little money as possible. However, he knows that there are always more people wanting to attend a concert than the number of seats in the Philharmonic hall. So he will not buy more than one ticket for one concert (separately or in a subscription), even if it allows him to save money.
Input
The first line contains integers n and m that are the number of concerts Denis wants to attend, and a number of different subscription types at the Philharmonic (2 ≤ n ≤ 10 5; 1 ≤ m ≤ 10 5). The i-th of the following n lines contains integers si and di that are the price of a ticket and the discount for students in percent at the i-th concert (100 ≤ si ≤ 50 000; 0 ≤ di ≤ 100). The j-th of the following m lines contains integers kj and pj that are the minimum number of concerts that can be combined in the subscription of the j-th type, and a discount for it in percent (2 ≤ kj ≤ n; 1 ≤ pj ≤ 100).
Output
Output the minimum amount of money Denis has to spend to attend all the concerts. The answer should be given with an absolute or relative error not exceeding 10 −6.
Sample Input
| input | output |
|---|---|
6 2
500 0
700 0
300 0
400 0
500 50
800 0
5 10
6 15
|
2680.00
|
Notes
In the example, Denis needs to buy a subscription of the first type (with 10% discount) for all concerts, except the fifth one, and to buy a ticket for the fifth concert with a student discount of 50%.
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int maxn=100000+10;
struct node{
ll val;int id;
bool operator<(const node &a) const{
return this->val>a.val;
}
};
vector<int> G[105];
priority_queue<node> q;
int num[105],g[105];
int main()
{
int n,m;
while(~SC("%d%d",&n,&m)){
ll ori=0,ans=0;
for(int i=0;i<=100;i++) G[i].clear();
FOR(i,n) {
int cost,p;
SC("%d%d",&cost,&p);
ori+=cost*(100-p);
G[p].push_back(cost);
}
for(int i=0;i<=100;i++) sort(G[i].begin(),G[i].end());
MM(num,inf);
FOR(i,m){
int peo,dis;
SC("%d%d",&peo,&dis);
num[dis]=min(num[dis],peo);
}
// PF("111111
");
ans=ori;
FOR(k,100){
if(num[k]>n) CT;
MM(g,0);
// PF("2222222
");
ll tmp=ori,cnt=0;
for(int i=0;i<=k;i++)
for(int j=0;j<G[i].size();j++){
tmp+=G[i][j]*(i-k);cnt++;
}
// PF("3333333
");
while(q.size()) q.pop();
for(int i=k+1;i<=100;i++) {
if(G[i].size()) q.push((node){G[i][0]*(i-k),i});
g[i]=1;
}
// PF("444444
");
while(cnt<num[k]){
node u=q.top();q.pop();
tmp+=u.val;
cnt++;
if(g[u.id]<G[u.id].size()){
int r=g[u.id];
q.push((node){G[u.id][r]*(u.id-k),u.id});
g[u.id]++;
}
}
ans=min(ans,tmp);
}
PF("%.2f
",(double)ans/100.00);
}
return 0;
}
错点：如果直接枚举优惠券的优惠值大小k（0-100）,然后，计算出每种票在此优惠券下的可优惠
价格，再sort排下序是100*1e5*log1e5是会超时的，所以需要优化一下，考虑买单个票的时候，打的折
<=k的必须要算（可优惠），对于>k的，对每种k，每次都只能选其中的一种（贪心），所以，可用优先队列将复杂度降到1e5log100，
注意==k的也必须要贪啊，，因为要让大与k的数量尽可能小