ZOJ 3875 Lunch Time（结构体排序）

女爷i 2022-07-25 16:45 86阅读 0赞

Lunch Time

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Time Limit: 2 Seconds  Memory Limit: 65536 KB

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The 999th Zhejiang Provincial Collegiate Programming Contest will be held in Marjar University. The canteen of Marjar University is making preparations for this grand competition. The canteen provides a lunch set of three types: appetizer, main course and dessert. Each type has several dishes with different prices for choosing.

Edward is the headmaster of Marjar University. One day, to inspect the quality of dishes, he go to the canteen and decides to choose a**median set** for his lunch. That means he must choose one dish from each of appetizers, main courses and desserts. Each chosen dish should at the median price among all dishes of the same type.

For example, if there are five dessert dishes selling at the price of 2, 3, 5, 10, 30, Edward should choose the dish with price 5 as his dessert since its price is located at the median place of the dessert type. If the number of dishes of a type is even, Edward will choose the dish which is more expensive among the two medians.

You are given the list of all dishes, please write a program to help Edward decide which dishes he should choose.

#### Input ####

There are multiple test cases. The first line of input contains an integer  T indicating the number of test cases. For each test case:

The first line contains three integers S, M and D (1 <=S, M, D <= 100), which means that there are S dishes of appetizer, M dishes of main course and D dishes of dessert.

Then followed by three parts. The first part contains S lines, the second and the last part containsM and D lines respectively. In each line of the three parts, there is a string and an integer indicating the name and the price of a dish. The name of dishes will only consist of non-whitespace characters with no more than 50 characters. The price of dishes are non-negative integers less than or equal to 1000. All dish names will be distinct.

#### Output ####

For each test case, output the total price of the median set, together with the names of appetizer, main course and dessert, separated by a single space.

#### Sample Input ####

2
    1 3 2
    Fresh_Cucumber 4
    Chow_Mein 5
    Rice_Served_with_Duck_Leg 12
    Fried_Vermicelli 7
    Steamed_Dumpling 3
    Steamed_Stuffed_Bun 4
    2 3 1
    Stir-fried_Loofah_with_Dried_Bamboo_Shoot 33
    West_Lake_Water_Shield_Soup 36
    DongPo's_Braised_Pork 54
    West_Lake_Fish_in_Vinegar 48
    Longjing_Shrimp 188
    DongPo's_Crisp 18

#### Sample Output ####

15 Fresh_Cucumber Fried_Vermicelli Steamed_Stuffed_Bun
    108 West_Lake_Water_Shield_Soup DongPo's_Braised_Pork DongPo's_Crisp 
    
    
    
    题意是说，有S, M and D三种食物，分别有S, M and D个，拿食物的时候取中间价值的，如果没有刚好中间价值的就取较中间的价值中大的那个。输出拿的三种食物的总价值和名称。
    
    
    
     /* 
    * Copyright (c) 2016, 烟台大学计算机与控制工程学院 
    * All rights reserved. 
    * 文件名称：zoj.cpp 
    * 作    者：单昕昕 
    * 完成日期：2016年4月2日 
    * 版 本 号：v1.0 
    */  
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    struct N//结构体保存名称和价值
    {
        string na;
        int v;
    };
    int cmp(const void *x,const void *y)//结构体排序
    {
        return (((N*)x)->v>((N*)y)->v);//对价值降序排列
    }
    int main()
    {
        N a[1005];
        int t;
        cin>>t;
        int s,m,d,ans;
        string x,y,z;
        while(t--)
        {
            ans=0;
            cin>>s>>m>>d;
            int i;
            //s
            for(i=0; i<s; ++i)
            {
                cin>>a[i].na;
                cin>>a[i].v;
            }
            qsort(a,s,sizeof(a[0]),cmp);//排序
            ans+=(a[(s/2)].v);//取中间价值
            x=a[(s/2)].na;//记录名称
            //m
            for(i=0; i<m; ++i)
            {
                cin>>a[i].na;
                cin>>a[i].v;
            }
            qsort(a,m,sizeof(a[0]),cmp);
            ans+=(a[(m/2)].v);
            y=a[(m/2)].na;
            //d
            for(i=0; i<d; ++i)
            {
                cin>>a[i].na;
                cin>>a[i].v;
            }
            qsort(a,d,sizeof(a[0]),cmp);
            ans+=(a[(d/2)].v);
            z=a[(d/2)].na;
            cout<<ans<<" "<<x<<" "<<y<<" "<<z<<endl;
        }
        return 0;
    }