http://acm.hdu.edu.cn/showproblem.php?pid=1052

Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31270    Accepted Submission(s): 9523

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

用dp来做确实有点慢，用到了贪心的思维。首先有一个性质是假设每次齐王都派出最快的马来pk，那么田忌的最优策略一定是他的最快马或者最慢马，如果田输了这场比赛更优显然用最小的马输给齐王更好。如果田想赢了这场比赛显然应该用最快的马，因为如果第二快的马能赢的话使用顺序无可厚非如果不能的话那就输了还不是用的最慢的马。

有f[i][j]=max(f[i+1][j]+cmp(a[i],b[j-i+1]),f[i][j-1]+cmp(a[j],b[j-i+1])); f[i][j]表示田的马为i~j时当前pk的马是j-i+1时的最优解，答案为f[1][n];

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<vector>
 6 using namespace std;
 7 #define inf 0x3f3f3f3f
 8 int f[1005][1005];
 9 int a[1005],b[1005];
10 int cmp(int i,int j)
11 {
12     if(a[i]>b[j])return 1;
13     if(a[i]==b[j])return 0;
14     return -1;
15 }
16 int main()
17 {
18     int N,M,i,j,k;
19     while(cin>>N&&N){
20         for(i=1;i<=N;++i) scanf("%d",a+i);
21         for(i=1;i<=N;++i) scanf("%d",b+i);
22         memset(f,0,sizeof(f));
23         sort(a+1,a+1+N);
24         sort(b+1,b+1+N);
25         for(i=1;i<=N;++i) f[i][i]=cmp(i,1);
26         for(int len=2;len<=N;++len)
27         {
28             for(i=1,j=len;j<=N;++i,++j)
29             {
30                 f[i][j]=max(f[i+1][j]+cmp(i,j-i+1),f[i][j-1]+cmp(j,j-i+1));
31             }
32         }
33         printf("%d\n",f[1][N]*200);
34     }
35     return 0;
36 }