题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1052
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Here is a famous story in Chinese history.

"That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and others."

"Both of Tian and the king have three horses in different classes, namely, regular, plus, and super. The rule is to have three rounds in a match; each of the horses must be used in one round. The winner of a single round takes two hundred silver dollars from the loser."

"Being the most powerful man in the country, the king has so nice horses that in each class his horse is better than Tian's. As a result, each time the king takes six hundred silver dollars from Tian."

"Tian Ji was not happy about that, until he met Sun Bin, one of the most famous generals in Chinese history. Using a little trick due to Sun, Tian Ji brought home two hundred silver dollars and such a grace in the next match."

"It was a rather simple trick. Using his regular class horse race against the super class from the king, they will certainly lose that round. But then his plus beat the king's regular, and his super beat the king's plus. What a simple trick. And how do you think of Tian Ji, the high ranked official in China?"



Were Tian Ji lives in nowadays, he will certainly laugh at himself. Even more, were he sitting in the ACM contest right now, he may discover that the horse racing problem can be simply viewed as finding the maximum matching in a bipartite graph. Draw Tian's horses on one side, and the king's horses on the other. Whenever one of Tian's horses can beat one from the king, we draw an edge between them, meaning we wish to establish this pair. Then, the problem of winning as many rounds as possible is just to find the maximum matching in this graph. If there are ties, the problem becomes more complicated, he needs to assign weights 0, 1, or -1 to all the possible edges, and find a maximum weighted perfect matching...

However, the horse racing problem is a very special case of bipartite matching. The graph is decided by the speed of the horses --- a vertex of higher speed always beat a vertex of lower speed. In this case, the weighted bipartite matching algorithm is a too advanced tool to deal with the problem.

In this problem, you are asked to write a program to solve this special case of matching problem.

Input

The input consists of up to 50 test cases. Each case starts with a positive integer n (n <= 1000) on the first line, which is the number of horses on each side. The next n integers on the second line are the speeds of Tian’s horses. Then the next n integers on the third line are the speeds of the king’s horses. The input ends with a line that has a single 0 after the last test case.

Output

For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.

Sample Input

3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0

Sample Output

200
0
0

Problem solving report:

Description: 田忌赛马，很老套的一个故事。田忌和他的国王赛马，田忌每赢一场就增加200美金，规定每匹马都只能出场一次，求田忌能拿到的最好结果，即最优解。
Problem solving: 这道题就是用自己很小的损失来换取对手很大的损失，贪心策略。
1. 如果a的最慢马大于b的最慢马，则直接用a的最慢马与b的最慢马比赛，赢一场；
2. 如果a的最慢马小于b的最慢马，则用a的最慢马浪费掉b的最快马，输一场；
3. 如果a的最慢速度等于b的最慢，则：
    ① 如果a的最快马大于b的最快马，则直接用a的最快马与b的最快马进行比赛，赢一场；
    ② 如果a的最快马小于b的最快马，则用a的最慢马浪费b掉的最快马，输一场；
    ③ 如果a的最快马等于b的最快马，即a与b的最慢与最快分别相等，则：
        Ⅰ 如果a的最慢马小于b的最快马，则用a的最慢马浪费掉b的最快马，输一场；
        Ⅱ 否则就是a的最快最慢马、b的最快最慢马都相等，说明剩余未比赛的马速度全部相等，直接结束比赛。

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int n, ans, al, ar,bl, br, a[1010], b[1010];
    while (cin >> n, n)
    {
        ans = 0;
        for (int i = 0; i < n; i++)
            cin >> a[i];
        for (int i = 0; i < n; i++)
            cin >> b[i];
        sort(a, a + n);
        sort(b, b + n);
        al = bl = 0;
        ar = br = n - 1;
        while (al <= ar && bl <= br)
        {
            if (a[al] > b[bl])
            {
                al++;
                bl++;
                ans += 200;
            }
            else if (a[al] < b[bl])
            {
                al++;
                br--;
                ans -= 200;
            }
            else
            {
                if (a[ar] > b[br])
                {
                    ar--;
                    br--;
                    ans += 200;
                }
                else if (a[ar] < b[br])
                {
                    al++;
                    br--;
                    ans -= 200;
                }
                else
                {
                    if(a[al] < b[br])
                    {
                        al++;
                        br--;
                        ans -= 200;
                    }
                    else break;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}