Problem C

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?"



C" title="Problem C">



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

题意：田忌赛马的超级版，田忌和大王赛马，但是相同等级的马谁的能跑还不一定，谁赢就能赢得对方200金币；

解题思路：这道题说是贪心还不如说成一道数学的分类讨论题，就和中学学的分类讨论太像了

1，田忌最快的马要是比大王最快的快，就比一句，田忌胜；

2，田忌最快的马比大王最快的马慢，就用田忌当前马中最慢的马和他比，田忌输；

3，田忌最快的马和大王最快的马速度相同，就要讨论最慢的马；

4，田忌的最慢的马比大王的马快，田忌胜；

5，否则，田忌输；

感悟：妈呀，这几天平均一天一道题的速度确实有点慢哈......马上就到星期天了，就能加快速了；

AC代码（G++
46MS）

#include

#include

#include

#include

#define maxn 1005

using namespace std;

struct  horses

{

    int
speed;

    char
s;

};

bool comp(horses &a,horses&b)

{

   
if(a.speed!=b.speed)

       
return a.speed>b.speed;

    else

       
return a.speed>b.speed;

}

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
n,money=0,tb,te,gb,ge;

    horses
ti[maxn],god[maxn];

   
while(~scanf("%d",&n)&&n)

    {

       
money=0;

       
tb=gb=0;

       
te=ge=n-1;

       
for(int i=0;i

           
scanf("%d",&ti[i].speed);

       
for(int i=0;i

           
scanf("%d",&god[i].speed);

       
sort(&ti[0],&ti[n],comp);

       
sort(&god[0],&god[n],comp);

       
while (tb <= te)

       
{

           
if (ti[te].speed > god[ge].speed)//田忌最快的马比大王的快就赢一局

           
{

               
money++;

               
te--;

               
ge--;

           
}

           
else if (ti[te].speed < god[ge].speed)//田忌最快的马比大王的慢就输一局

           
{

               
money--;

               
te--;

               
gb++;

           
}

           
else//最快的马一样快

           
{

               
if(ti[tb].speed > god[gb].speed)//田忌最慢的马比大王的快就赢一局

               
{

                   
money++;

                   
tb++;

                   
gb++;

               
}

               
else

               
{

                   
if (ti[te].speed < god[gb].speed)//田忌的慢就输

                   
money--;

                   
te--;

                   
gb++;

               
}

           
}

           
//注意：这些最快最慢都是相对于当前还没有进行过比赛的马，每比一次状态转移

           
//一次

      
}

       
printf("%d\n",money*200);

    }

    return
0;

}