SDAU课程练习1002

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each input case, output a line
containing a single number, which is the maximum money Tian Ji will
get, in silver dollars.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

200
0
0

题目大意：

田忌赛马，然后win一次 +200 lose 一次 -200.尽可能让田忌 win 输出最后的 金子数。

思路：

一开始是从快马开始考虑的，想着从快吗开始检索，找出最多的 win 马。

但是结果是错误的。

这个题应该分这几种去讨论：
1. 田忌慢马
> 齐王慢马 win ++；
2. 田忌慢马
< 齐王慢马 lose ++ ，齐王快马 out；
3. 田忌慢马 =
齐王慢马
{

          if（田忌快马
> 齐王快马） win ++ ；

          
else lose ++ 齐王快马 out；
}

然后开始模拟就行啦。

感想：

快发疯了，写上面的思路的时候  句子结束都不用 。 而是用 ；
如果也直接写的  if  关键是第三条 讨论还加了一个
花括号。。。。完了，我不想进化成  小猿猴啊。。。

AC代码：

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

using namespace std;

bool cmp(int a,int b)

{

    return
a>b;

}

int main()

{

   
//freopen("r.txt", "r", stdin);

    int
n,i,j,win,lost;

    int
a[5000];

    int
b[5000];

    int
t[5000];

   
while(cin>>n)

    {

       
if(n==0) break;

       
for(i=0;i

       
{

           
cin>>a[i];

       
}

       
for(i=0;i

       
{

           
cin>>b[i];

       
}

       
sort(a,a+n,cmp);

       
sort(b,b+n,cmp);

       
win=lost=0;

       
int tmax,tmin,kmax,kmin;

       
tmax=kmax=0;tmin=kmin=n-1;

       
while(tmax<=tmin)

       
{

           
if(a[tmin]>b[kmin])

           
{

               
win++;

               
tmin--;

               
kmin--;

           
}

           
else if(a[tmin]

           
{