Popular Products_太阁竞赛B

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本题旨在找出一组顾客购买清单中所有出现的商品。每个购买记录包括产品ID、购买日期及价格，相同ID但不同价格的商品被视为不同。需要处理多个购买清单，并输出所有清单中共有的商品ID。

题目2 : Popular Products
时间限制:10000ms
单点时限:1000ms
内存限制:256MB
描述
Given N lists of customer purchase, your task is to find the products that appear in all of the lists.

A purchase list consists of several lines. Each line contains 3 parts: the product id (format XXXX-XXXX), the purchase date (format mm/dd/yyyy) and the price (with decimal places). Two product are considered equal if both the product id and the price are equal.

输入
The first line contains an integer N denoting the number of lists. (1 ≤ N ≤ 1000)

Then follow N blocks. Each block describes one list.

The first line of each block contains an integer M denoting the number of products in the list. (1 ≤ M ≤ 50)

M lines follow. Each line contains the product id, the purchase date and the price.

输出
The products that appear in all of the lists. You should output the product id in increasing order.

If two different product share the same id (with different price) you should output the id twice.

样例输入
3
2
1111-1111 07/23/2016 998.00
1111-2222 07/23/2016 888.00
2
1111-2222 07/23/2016 888.00
1111-1111 07/23/2016 998.00
2
1111-2222 07/23/2016 888.00
1111-1111 07/23/2016 999.00
样例输出

1111-2222

【我的程序】

#include <iostream>
#include <string>
#define maxSize 51
using namespace std;

int main()
{
    int n, m, x, flag[maxSize]={0};
    string id, date, price, temp, refPd[maxSize], ls[maxSize];

    cin>>n >>m;
    for (int i=0;i<m;i++){
        cin>>id >>date >>price;
        refPd[i]=id+price;
    }

    for (int i=0;i<m-1;i++)
        for (int j=0;j<m-1-i;j++)
            if (refPd[j]>refPd[j+1]){ temp=refPd[j]; refPd[j]=refPd[j+1]; refPd[j+1]=temp; }
    for (int i=1;i<m;i++) if (refPd[i]==refPd[i-1]) flag[i]=-1;

    for (int i=1;i<n;i++){
        cin>>x;
        for (int j=0;j<x;j++){
            cin>>id >>date >>price;
            ls[j]=id+price;
        }
        for (int j=0;j<m;j++)
            if (flag[j]>=0) for (int k=0;k<x;k++) if (refPd[j]==ls[k]) { flag[j]++; break; }
    }

    for (int i=0;i<m;i++) if (flag[i]==n-1) cout<<refPd[i].substr(0,9)<<endl;
    return 0;
}