本文通过分析田忌赛马问题，提出了一种贪心算法解决方案。通过对不同情况进行讨论和证明，确保了算法的正确性和最优性。
"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.
92 83 71
95 87 74
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20 19
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感慨：（怀念小学啊）对于本题，我是真心的被误导了，网上流传用DP来解决，但是还是没做出来，最后还是用贪心做出来的。不过我最想说的是，大过年，这么冷，刷题不容易啊，这么难的题，有一个地方没考虑到根本就做不出
思路：因为是国王先出马，所以就要考虑分配马的情况了。
也许很多人会想，直接让田忌用最慢的马跟国王最快的马比，但是编了代码后会发现，如果出现
4 3 3
5 4 4
的情况这么弄明显不行了，我直接按常规出牌我会赚600金币，但是我按刚刚的思路就只会赚400。想清楚这一点，算法就可以建立了，我可以想出以下几种情况了：
一、当田忌最快的马比国王最快的马快时，用田忌最快的马赢国王最快的马。
二、当田忌最快的马比国王最快的马慢时，用田忌最慢的马输给国王最快的马。
三、当田忌最快的马跟国王最快的马一样快时，分情况（最难，真的太难考虑了百度后才知道最终的算法与考虑的情况）
证明一、：假设现在国王最快的马是K，田忌最快的马是T，如果存在一种更优的比赛策略，让T的对手不是K，而使得田忌赢更多的钱的话，那么设此时K的对手是t，T的对手是k：( T>K &&T>t && K>k)
1、 若t>K，则有T>k，t>K。这个结果和T>K，t>k是相同的。
2、 若k<t≤K，则有T>k，t≤K。这个结果不如T>K，t>k来得优秀。
3、 若t≤k≤K，则有T>k，t≤K。这个结果和T>K，t≤k是相同的。
由此可知，交换各自对手后，一定不会使得结果变劣，那么假设是不成立的。
证明二、：因为田忌最快的马比国王最快的马慢，所以田忌所有的马都比国王最快的马慢，也就是说此时田忌的所有马都赢不了国王的马，而出于贪心的思想，应该保留相比之下更快的马，因此用最慢的马去输一定不会比用别的马去输来得劣。
其实上面这两个很容易就想得到，最难的是相等的时候。因为不可以直接让田忌最快的马跟国王最快的马打平，或者直接用最慢的马去输给国王最快的马。（存在反例）
1、如果选择全部打平，那么对于田忌 1 2 3 4，国王 1 2 3 4 ，这组数据，田忌什么黄金也得不到。但是如果选择 1->4, 4->3, 3->2, 2->1田忌可以得到400两黄金。
2、如果选择用最慢的马输掉比赛的话，对于田忌 3 4，国王 1 4 ，这组数据，田忌一胜一负，什么黄金也得不到，但是如果田忌选择 3->1 , 4->4 ,一胜一平，田忌可以得到200两黄金。
所以：对于情况三，我们应该从最慢的马开始考虑了
1、当田忌最慢的马比国王最慢的马快，那么用田忌最慢的马赢国王最慢的马
2、当田忌最慢的马比国王最慢的马慢，那么用田忌最慢的马输给国王最快的马
3、当田忌最慢的马跟国王最慢的马相等的时候，用田忌最慢的马跟国王最快的马比
证明1、：假设现在国王最慢的马是K，田忌最慢的马是T，如果存在一种更优的比赛策略，让T的对手不是K，而使得田忌赢更多的钱的话，那么设此时K的对手是t，T的对手是k：( T>K &&t>T && k>K)
1、 若T>k，则有T>k，t>K。这个结果和T>K，t>k是相同的。
2、 若T<k≤t，则有T<k，t>K。。这个结果不如T>K，t>k来得优秀。
3、 若K≤t≤k，则有T>k，t≤K。这个结果和T>K，t≤k是相同的.
由此可知，交换各自对手后，一定不会使得结果变劣，那么假设是不成立的。
证明2、：因为田忌最慢的马比国王最慢的马慢，所以田忌最慢的马都比国王所有的马慢，也就是说此时田忌最慢的马赢不了国王的任何一匹马，而出于贪心的思想，应该去掉国王最快的马，因此输给国王最快的马一定不会比输给其他的马来得劣。
证明3、：因为田忌最快T和最慢t的马与国王最快K和最慢k的马都相等，如果用最快的马跟国王最快的马比，最慢的马和国王最慢的马比，那么田忌什么收获也没有。此时是T->K, t->k,其实这个时候我们可以换个角度想，上面这种情况跟T->k, t->k,没有什么区别，都是双方少了最快和最慢的马。但是从证明二可知，我们必须让最慢的马输给国王最快的马，而在田忌的马中找任何一匹比最慢的马快一点的马去赢国王最慢的马都行，这样田忌在同样一胜一负的情况下，保留了跑得更快的马，会比把最快的马拿去比赛来得更优些。
所以代码如下：
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 1010
int T[N],K[N];
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int ti,tj,ki,kj,win,n;
while(scanf("%d",&n)!=EOF)
{
int i;
if(n==0)break;
for(i=0;i<n;i++)
scanf("%d",&T[i]);
for(i=0;i<n;i++)
scanf("%d",&K[i]);
sort(T,T+n,cmp);
sort(K,K+n,cmp);
ti=ki=0;
tj=kj=n-1;
for(i=0,win=0;i<n;i++)
{
if(T[ti]>K[ki])
{
ti++;
ki++;
win++;
}
else if(T[ti]<T[ki])
{
tj--;
ki++;
win--;
}
else
{
if(T[tj]>K[kj])
{
tj--;
kj--;
win++;
}
else if(T[tj]<K[kj])
{
tj--;
ki++;
win--;
}
else
{
if(T[tj]<K[ki])
{
tj--;
ki++;
win--;
}
}
}
}
printf("%d\n",win*200);
}
return 0;
}
感言：ACM真的需要坚持，好几次想放弃这道题，但是觉得还是来搞一下。希望以后能真正的独挡一面吧！
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