CodeForces 416C:Booking System(贪心)
2017-08-06 15:44
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Booking System
Time limit:1000 ms Memory limit:262144 kBProblem Description
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.
Example
Input |
---|
3 |
10 50 |
2 100 |
5 30 |
3 |
4 6 9 |
Output |
2 130 |
2 1 |
3 2 |
Note
In the first sample, Genos can destroy the entire line in one second. In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds. In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
题意:
点菜系统,同一时间有n批顾客,每批顾客有 Ci 个人,并且消费 Pi 金额。餐厅有 k 个桌子,每个桌子能坐下 Ri 个人,每个桌子只能坐一批客人,问如何安排,才能使得消费金额最高。解题思路:
由于每个桌子只能坐一批客人,对于每个桌子来说应尽可能的选择消费金额高的顾客。贪心策略:将消费金额从大到小排序,桌子容量从小到大排序,先安排消费金额最高的客人。对于每批能安排的客人,我们找最小的能够安排它的桌子来安排。
Code:
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int maxn=1000; struct Node { int no; int c;//人数 int p;//钱 }a[maxn]; struct Node1 { int no; int r; int vis; }b[maxn]; struct Node2 { int a; int b; }ans[maxn]; bool cmp(Node i,Node j) { if(i.p==j.p) return i.c<j.c; return i.p>j.p; } bool cmp1(Node1 i,Node1 j) { return i.r<j.r; } int main() { int n,k; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&a[i].c,&a[i].p); a[i].no=i+1; } scanf("%d",&k); for(int i=0;i<k;i++) { scanf("%d",&b[i].r); b[i].no=i+1; b[i].vis=0; } sort(a,a+n,cmp); sort(b,b+k,cmp1); int cnt=0,sum=0; for(int i=0;i<n;i++) { for(int j=0;j<k;j++) { if(b[j].vis==0&&b[j].r>=a[i].c) { b[j].vis=1; sum+=a[i].p; ans[cnt].a=a[i].no; ans[cnt].b=b[j].no; cnt++; break; } } } printf("%d %d\n",cnt,sum); for(int i=0;i<cnt;i++) { printf("%d %d\n",ans[i].a,ans[i].b); } return 0; }
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