UVa1344 - Tian Ji -- The Horse Racing

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 ( n1000) 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
题目大意：田忌和齐王赛马，田忌和齐王各有N匹马。输了赔200两银子，赢了赚200两银子。现在已知每匹马的速度，问田忌最多能够赚多少两银子。
题解：很考察人的细心程度，各方面都要考虑周全，被if else搞晕了+_+。。。。我们先对田忌和齐王的马排序，接下来就进行匹配了。数组a表示田忌的马的速度，数组b表示齐王的马的速度，ansa，ansb分别为田忌和齐王未进行匹配的速度最大的马，cura,curb为田忌和齐王未进行匹配的速度最小的马。
我们采取的贪心策略是：如果a[ansa]>b[ansb]则匹配成功，并赚200两，如果a[ansa]<b[ansb]则我们拿当前最垃圾的马（a[cura])和齐王的最好的马（b[ansb])进行匹配，并赔200两，如果相等（a[ansa]==b[ansb])则再进一步比较，如果a[cura]>b[curb]则进行匹配，并赚两百两，如果a[cura]<b[curb]则拿最垃圾的马（a[cura]）和齐王最好的马（b[ansb])进行匹配，并赔两百两，如果a[cura]==b[curb]则再判断一下，如果a[cura]<b[ansb]则两者进行匹配，并赔200两，否则的话就真的只能打平手了。真心是蛋碎一地啊!!!....

View Code

 1 #include<stdio.h>

 2 #include<stdlib.h>

 3 #include<string.h>

 4 #define MAXN 1005

 5 long a[MAXN],b[MAXN],f[MAXN],v[MAXN];

 6 int compare(const void*a,const void*b)

 7 {

 8     return *(int*)a-*(int*)b;

 9 }

10 int main(void)

11 {

12     long n,i,cura,curb,ans,ansa,ansb;

13     while(scanf("%ld",&n)!=EOF)

14     {

15         if(n==0) break;

16         for(i=0; i<n; i++)

17             scanf("%ld",&a[i]);

18         for(i=0; i<n; i++)

19             scanf("%ld",&b[i]);

20         qsort(a,n,sizeof(a[0]),compare);

21         qsort(b,n,sizeof(b[0]),compare);

22         memset(f,0,sizeof(f));

23         ans=0;

24         cura=0;

25         curb=0;

26         ansa=n-1;

27         ansb=n-1;

28         while(n--)

29         {

30             if(a[ansa]>b[ansb])

31             {

32                 ans++;

33                 ansa--;

34                 ansb--;

35             }

36             else

37             if(a[ansa]<b[ansb])

38             {

39                 ans--;

40                 cura++;

41                 ansb--;

42             }

43             else

44             {

45                 if(a[cura]<b[curb])

46                 {

47                     ans--;

48                     cura++;

49                     ansb--;

50                 }

51                 else

52                 if(a[cura]>b[curb])

53                 {

54                     ans++;

55                     cura++;

56                     curb++;

57                 }

58                 else

59                 if(a[cura]<b[ansb])

60                 {

61                     ans--;

62                     cura++;

63                     ansb--;

64                 }

65             }

66         }

67         printf("%ld\n",ans*200);

68     }

69     return 0;

70 }