codeforces 416C C. Booking System(贪心)-优快云博客

本文讨论了使用贪心算法解决餐厅预订系统优化的问题，通过合理分配桌子和预订请求，最大化收入。具体实现包括对预订请求按价值排序，然后尝试为每个请求分配合适的桌子，确保满足最大桌容量限制，最终输出最优的接受请求数量和总收益。

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1、http://codeforces.com/problemset/problem/416/C

2、思路：

贪心题目

按照钱数从大到小排序，钱数相同按照人数从小到大排序，然后看是否有合适的椅子分配给这个团队，如果可以，就记录下来，输出即可

3、题目：

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:c_i andp_i — 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 c_i 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 knowr_i — 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. Thenn lines follow. Each line contains two integers:c_i, p_i(1 ≤ c_i, p_i ≤ 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 containsk space-separated integers: r₁, r₂, ..., r_k(1 ≤ r_i ≤ 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.

Sample test(s)

Input

Output

2 130
2 1
3 2

3、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int w;
    int v;
    int id;
} a[1005];
int m;
struct node1
{
    int k;
    int id;
}b[1005];
int cmp(node b,node c)
{
    if(b.v==c.v)
        return b.w<c.w;
    return b.v>c.v;
}
int cmp1(node1 b,node1 c)
{
    return b.k<c.k;
}
int find(int p)
{
    for(int i=1;i<=m;i++)
    {
        if(b[i].k!=-1 && b[i].k>=p)
        {
            b[i].k=-1;
            return b[i].id;
        }
    }
    return 0;
}
int main()
{
    int n;
    memset(b,-1,sizeof(b));
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%d%d",&a[i].w,&a[i].v);
        a[i].id=i;
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&b[i].k);
        b[i].id=i;
    }
    sort(b+1,b+m+1,cmp1);
    sort(a+1,a+n+1,cmp);
    node1 c[1005];
    int j=1;
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        int tmp=find(a[i].w);
        if(tmp)
        {
            ans+=a[i].v;
            c[j].k=a[i].id;
            c[j].id=tmp;
            j++;
        }
    }
    printf("%d %d\n",j-1,ans);
    for(int i=1;i<j;i++)
    {
        printf("%d %d\n",c[i].k,c[i].id);
    }
    return 0;
}