Shopping

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6348 Accepted Submission(s): 2215

Problem Description

Every girl likes shopping,so does dandelion.Now she finds the shop is increasing the price every day because the Spring Festival is coming .She is fond of a shop which is called "memory". Now she wants to know the rank of this shop's price after the change of everyday.

Input

One line contians a number n ( n<=10000),stands for the number of shops.
Then n lines ,each line contains a string (the length is short than 31 and only contains lowercase letters and capital letters.)stands for the name of the shop.
Then a line contians a number m (1<=m<=50),stands for the days .
Then m parts , every parts contians n lines , each line contians a number s and a string p ,stands for this day ,the shop p 's price has increased s.

Output

Contains m lines ,In the ith line print a number of the shop "memory" 's rank after the ith day. We define the rank as :If there are t shops' price is higher than the "memory" , than its rank is t+1.

Sample Input

memory

kfc

wind

49 memory

49 kfc

48 wind

80 kfc

85 wind

83 memory

Sample Output

理解题意：

这个题的题意很明确，让输出“memory”这个商店每天的价格排名

解题思路：

首先，map是关联容器，他可以实现从键值到值的映射，类似函数中的一一对应；然后我们定义一个map<string,int>shop;用来存储商店的名字和每天的价格；注意给每个商店赋值价格的操作是shop[s] += p;但是如果用下面这种方法进行排名的话，时间就会超限

            int rank = 1;
            map<string,int>::iterator it;
            for(it = shop.begin();it != shop.end(); it++){
                if(it->second>shop["memory"])
                    rank++;
            }
            cout<<rank<<endl;

所以我经过查阅资料发现了另一种方法，用一个数组来把价格存起来，然后用sort排序后，再通过比较数组中的值是否和“memory”商店的价格相等来输出他的位置，即为他的排名

#include <iostream>
#include <string>
#include <map>
#include <algorithm>

using namespace std;

bool my_camp(int x,int y){
    return x>y;
}

int main()
{
    int n,m,p;
    int a[10001];
    map<string,int>shop;
    while(cin>>n){
        string s;
        for(int i = 1;i <= n; i++)
            cin>>s;
        cin>>m;
        while(m--){
            for(int i = 1;i <= n; i++){
                cin>>p>>s;
                shop[s] += p;
                a[i] = shop[s];
            }
            /*int rank = 1;
            map<string,int>::iterator it;
            for(it = shop.begin();it != shop.end(); it++){
                if(it->second>shop["memory"])
                    rank++;
            }
            cout<<rank<<endl; */
            sort(a+1,a+n+1,my_camp);
            for(int i = 1;i <= n; i++){
                if(a[i] == shop["memory"]){
                    cout<<i<<endl;
                    break;
                }
            }
        }
        shop.clear();
    } 
    return 0;
}

来源：https://www.cnblogs.com/jingshixin/p/12241794.html

标签

memory

map的使用-Hdu 2648

Shopping