11136-Hoax or what

Each Mal-Wart supermarket has prepared a promotion scheme run by the following rules:

• A client who wants to participate in the promotion (aka a sucker) must write down their phone number on the bill of their purchase and put the bill into a special urn.
• Two bills are selected from the urn at the end of each day: first the highest bill is selected and then the lowest bill is selected. The client who paid the largest bill receives a monetary prize equal to the difference between his bill and the lowest bill of the day.
• Both selected bills are not returned to the urn while all the remaining ones are kept in the urn for the next day.
• Mal-Wart has many clients such that at the end of each day there are at least two bills in the urn.
• It is quite obvious why Mal-Wart is doing this: they sell crappy products which break quickly and irreparably. They give a short-term warranty on their products but in order to obtain a warranty replacement you need the bill of sale. So if you are gullible enough to participate in the promotion you will regret it.

Your task is to write a program which takes information about the bills put into the urn and computes Mal-Wart's cost of the promotion.

The input contains a number of cases. The first line in each case contains an integer n, 1<=n<=5000, the number of days of the promotion. Each of the subsequent n lines contains a sequence of non-negative integers separated by whitespace. The numbers in the (i+1)-st line of a case give the data for the i-th day. The first number in each of these lines, k, 0≤k≤10⁵, is the number of bills and the subsequent k numbers are positive integers of the bill amounts. No bill is bigger than 10⁶. The total number of all bills is no bigger than 10⁶. The case when n = 0 terminates the input and should not be processed.

For each case of input print one number: the total amount paid to the clients by Mal-Wart as the result of the promotion.

Sample input

5
3 1 2 3
2 1 1
4 10 5 5 1
0
1 2
2
2 1 2
2 1 2
0

Output for sample input

19
2



动态数组最大值最小值差值的和。Multiset。

#include<cstdio>
#include<set>
using namespace std;
typedef long long LL;
int main()
{
    LL n,x,maxn,mind,d,sum;
    multiset<LL> s;
    while(scanf("%I64d",&n)&&n!=0)
    {
        sum=0;
        s.clear();
        for(LL i=0; i<n; i++)
        {
            scanf("%I64d",&x);
            for(LL j=0; j<x; j++)
            {
                scanf("%I64d",&d);
                s.insert(d);
            }
            multiset<LL>::iterator it;
            it=s.begin();
            mind=*it;
            s.erase(it);
            it=s.end(),it--;
            maxn=*it;
            s.erase(it);
            sum+=maxn-mind;
        }
        printf("%lld\n",sum);
    }
    return 0;
}