• 订票


    题目网址: http://codeforces.com/problemset/problem/416/C

    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.

    Sample Input

    Input
    3
    10 50
    2 100
    5 30
    3
    4 6 9
    Output
    2 130
    2 1
    3 2

    思路:先将钱数由大到小排序,若钱相等就按人数 人少排在前面(注意记录来的人的顺序,记录),将桌子的容客量由小到大排序。
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    using namespace std;
    int ss[1005][2];
    
    struct data
    {
        int c,p,i;
    }a[1005];
    
    struct hh
    {
        int rr,i;
    }r[1005];
    
    bool cmp(data a,data b)
    {
        if(a.p==b.p)
        return a.c<b.c;
        return a.p>b.p;
    }
    
    bool cmm(hh a,hh b)
    {
        return a.rr<b.rr;
    }
    
    int main()
    {
        int n,k;
    
        while(cin>>n)
      {
        for(int i=0;i<n;i++)
        {
            cin>>a[i].c>>a[i].p;
            a[i].i=i+1;
        }
        cin>>k;
        for(int i=0;i<k;i++)
        {
            cin>>r[i].rr;
            r[i].i=i+1;
        }
        sort(a,a+n,cmp);
        sort(r,r+k,cmm);
        int num=0,ans=0;
        for(int i=0;i<n;i++)
        for(int j=0;j<k;j++)   //钱数多的优先找桌子,且找能容纳这群人的座位数最少的;
        if(a[i].c<=r[j].rr)
        {
            ans+=a[i].p;
            r[j].rr=0;
            ss[num][0]=a[i].i;
            ss[num++][1]=r[j].i;
            break;
        }
        cout<<num<<' '<<ans<<endl;
        for(int i=0;i<num;i++)
        cout<<ss[i][0]<<' '<<ss[i][1]<<endl;
      }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chen9510/p/5022282.html
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