## 题目描述

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.

## 思路

1. Tian Ji最快的马比King最快的马快，直接赢。
2. Tian Ji最快的马与King最快的马速度一样，则
1） 拿Tian Ji最慢的马与King的马比，速度比King大则赢掉
2） 否则，拿Tian Ji最慢的马与King最快的马比，若速度一样则平局（可能出现速度都一样的情况），若速度小则输掉King最大速度的马
3. Tian Ji最快的马比King最快的马慢，则拿Tian Ji最慢的马输掉King最快的马。

## AC代码

cpp
/*

File Name: 1052.cpp
Author: Netcan
Blog: http://www.netcan.xyz
Mail: 1469709759@qq.com
Created Time: Fri 05 Jun 2015 07:50:23 PM CST
**/

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include
using namespace std;

int main()
{

#ifdef Oj
freopen(“1052.in”, “r”, stdin);

#endif
int n;
vector tian, king;
while(cin >> n && n) {
int x;
for (int i = 0; i < n; ++i) {
cin >> x;
tian.push_back(x);
}
for (int i = 0; i < n; ++i) {
cin >> x;
king.push_back(x);
}
sort(tian.begin(), tian.end());
sort(king.begin(), king.end());
int ti, tj, ki, kj; // Tian ji最慢、最快的马，King最慢最快的马
ti = ki = 0;
tj = kj = n-1;
int fee = 0;
for (int i = n-1; i >=0; –i) {
if(tian[tj] > king[kj]) { // 情况1
fee+=200;
–tj, –kj;
}
else if(tian[tj] == king[kj]) { // 情况2
if(tian[ti] > king[ki]) {
fee+=200;
++ti, ++ki;
}
else if(tian[ti] <= king[ki]) {
if(tian[ti] < king[kj]) {
fee-=200;
++ti, –kj;
}
else if(tian[ti] == king[kj]) {
++ti, –kj;
}
}
}
else if(tian[tj] < king[kj]) { // 情况3
fee-=200;
++ti, –kj;
}
}
cout << fee << endl;

    tian.clear();
king.clear();
}

return 0;


}