Hdu 1052 Tian Ji -- The Horse Racing

Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18291    Accepted Submission(s):
5327

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?"

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

 

 贪心题：田忌赛马----田忌和齐王各有n匹马，每匹马都得比赛，赢一局赢200，输一局输200，问怎么比才能使田忌赢得的金钱最多？

 

要想赢得最多，田忌应该采取避其锋芒的策略，首先将二者的所有马匹按从大到小的顺序排列，然后采取以下规则：

如果田忌最快的马比齐王最快的马快，则pk掉，田忌赢一局；

如果田忌最快的马比齐王最快的马慢，则用田忌最慢的马pk齐王最快的马，田忌输一局；

否则的话，

　　如果田忌最慢的马比齐王最慢的马快，则pk掉，田忌赢一局；

　　如果田忌最慢的马比齐王最慢的马慢，则用田忌最慢的马pk齐王最快的马，田忌输一局；

　　如果田忌最快的马与最慢马和齐王最快的马与最慢的马都相等，则用田忌最慢的马pk齐王最快的马。

 

 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define N 1005
 int tian[N],king[N];

 int cmp(int a,int b)
 {
     return a>b;
 }
 int main()
 {
     int n, i, j;
     int counts, i1, j1;
     while(cin>>n && n)
     {
         for(i=; i<n; i++)
             cin>>tian[i];
         for(i=; i<n; i++)
             cin>>king[i];
         sort(tian,tian+n,cmp);
         sort(king,king+n,cmp);
         counts = ;
         i = i1 = ;
         j = j1 = n-;
         while(i<=j)
         {
             if(tian[i]>king[i1])        //田忌最快的马比齐王最快的马快
             {
                 counts++;
                 i++;
                 i1++;
             }
             else if(tian[i]<king[i1])   //田忌最快的马比齐王最快的马慢
             {
                 counts--;
                 j--;
                 i1++;
             }
             else
             {
                 if(tian[j]>king[j1])    //田忌最慢的马比齐王最慢的马快
                 {
                     counts++;
                     j--;
                     j1--;
                 }
                 else if(tian[j]<king[j1])   //田忌最慢的马比齐王最慢的马慢
                 {
                     counts--;
                     j--;
                     i1++;
                 }
                 else
                 {
                     if(tian[j] != king[i1]) //田忌最慢的马和齐王最快的马不相等
                         counts--;
                     j--;
                     i1++;
                 }
             }
         }
         cout<<counts*<<endl;
     }
     return ;
 }