Popular Products_太閣競賽B

題目2 : Popular Products
時間限制:10000ms
單點時限:1000ms
內存限制: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.  


輸入
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.  


輸出
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.  


樣例輸入
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  
樣例輸出

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;
}