Booking System CodeForces - 416C
Innovation technologies are on a victorious march around the planet. They integrate into all spheres of human activity!
A restaurant called “Dijkstra’s Place” has started thinking about optimizing the booking system.
There are n booking requests received by now. Each request is characterized by two numbers: ci and pi — the size of the group of visitors who will come via this request and the total sum of money they will spend in the restaurant, correspondingly.
We know that for each request, all ci people want to sit at the same table and are going to spend the whole evening in the restaurant, from the opening moment at 18:00 to the closing moment.
Unfortunately, there only are k tables in the restaurant. For each table, we know ri — the maximum number of people who can sit at it. A table can have only people from the same group sitting at it. If you cannot find a large enough table for the whole group, then all visitors leave and naturally, pay nothing.
Your task is: given the tables and the requests, decide which requests to accept and which requests to decline so that the money paid by the happy and full visitors was maximum.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of requests from visitors. Then n lines follow. Each line contains two integers: ci, pi (1 ≤ ci, pi ≤ 1000) — the size of the group of visitors who will come by the i-th request and the total sum of money they will pay when they visit the restaurant, correspondingly.
The next line contains integer k (1 ≤ k ≤ 1000) — the number of tables in the restaurant. The last line contains k space-separated integers: r1, r2, …, rk (1 ≤ ri ≤ 1000) — the maximum number of people that can sit at each table.
Output
In the first line print two integers: m, s — the number of accepted requests and the total money you get from these requests, correspondingly.
Then print m lines — each line must contain two space-separated integers: the number of the accepted request and the number of the table to seat people who come via this request. The requests and the tables are consecutively numbered starting from 1 in the order in which they are given in the input.
If there are multiple optimal answers, print any of them.
Examples
Input:
3
10 50
2 100
5 30
3
4 6 9
Output:
2 130
2 1
3 2
思路：
有n个订单和k张桌子，要求利益最大，需要输出最大的利益和保留的订单对应的桌号。订单需要一个结构体，
需要二级排序，钱是从大到小排，人数从小到排。桌子我是从小到大排的，但好像没啥必要，反正订单和桌子
都匹配了一遍，下面是ac代码：
#include<bits/stdc++.h>
using namespace std;
struct node
{
int peo,mon,hao;
bool operator<(const node &a)const{
return a.mon==mon?a.peo>peo:a.mon<mon;
}
}req[1005];
struct node1
{
int re;
int ta;
}a[10005];
struct node2
{
int tabb;
int hao;
bool operator<(const node2 &a)const{
return a.tabb>tabb;
}
}tab[1005];
int main()
{
int n,k,rmax=0,mmax=0;
int book[1005];
memset(book,0,sizeof(book));
cin>>n;
for(int i=0;i<n;i++){
cin>>req[i].peo>>req[i].mon;
req[i].hao=i+1;
}
cin>>k;
for(int i=0;i<k;i++){
cin>>tab[i].tabb;
tab[i].hao=i+1;
}
sort(req,req+n);
sort(tab,tab+k);
/*for(int i=0;i<n;i++){
cout<<req[i].peo<<" "<<req[i].mon<<endl;
}
for(int i=0;i<k;i++){
cout<<tab[i].tabb<" ";
}
cout<<endl;*/
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
//cout<<req[i].peo<<" "<<tab[j].tabb<<endl;
if(req[i].peo<=tab[j].tabb&&book[j]==0){
//cout<<1<<endl;
a[rmax].re=req[i].hao;
a[rmax].ta=tab[j].hao;
rmax++;
mmax+=req[i].mon;
book[j]=1;
break;
}
}
}
cout<<rmax<<" "<<mmax<<endl;
for(int i=0;i<rmax;i++){
cout<<a[i].re<<" "<<a[i].ta<<endl;
}
}