NYOJ 364 田忌赛马

田忌赛马

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

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.

输入

The input consists of many 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.

输出

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

样例输入

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

样例输出

200
0
0

田忌赛马，不解释了。。。中国人都知道

算法分析：

刚开始想着直接从起点开始比较的，但遇到两个数相同甚至后面连续多个相同的的情况时考虑起来就复杂了，对于相同的两个马，比还是不比是问题的思考关键，测试解决了

两种特殊的情况比如：95 90 88 76 68 and 99 95 90 88 60以及95 95 90 80 and 99 95 95 95,都是对的，但测试了下WA，肯定还是哪里考虑不周全，转换一下思路：

1、当田忌最快的马比齐王最快的马快时，用田忌最快的马对抗齐王最快的马，赢一场。

2、当田忌最慢的马比齐王最慢的马快时，用田忌最慢的马对抗齐王最慢的马，赢一场。

3、当1、2都不满足时，用田忌最慢的马对抗齐王最快的马，若田忌最慢的马比齐王最快的慢时，输一场，否则平局。

注意:

if(a[tr]<b[kl])  //剔除相邻四个数全部相等那一个情况
                ans--;  

完整代码：

#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
const int MAXN = 1000;
int a[MAXN+5],b[MAXN+5];
int match(int n)
{
	int tl=0,tr=n-1,kl=0,kr=n-1,ans=0;
	while(tl<=tr)
	{
		if(a[tl]>b[kl])
		{
			ans++;tl++;kl++;
		}
		else if(a[tr]>b[kr])
		{
			ans++;tr--;kr--;
		}
		else
		{
			if(a[tr]<b[kl])  //剔除相邻四个数全部相等那一个情况
                ans--;  
			tr--;kl++;
		}
	}
	return ans*200;
}
int main()
{
	int n,i;
	while(cin>>n)
	{
		for(i=0;i<n;i++)
			cin>>a[i];
		for(i=0;i<n;i++)
			cin>>b[i];
		sort(a,a+n,greater<int>());
		sort(b,b+n,greater<int>());
		
		cout<<match(n)<<endl;
	}
	return 0;
}