HDU 1052 Tian Ji -- The Horse Racing 贪心

Tian Ji – The Horse Racing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24176 Accepted Submission(s): 7090

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

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

这个题真心不好想，个人感觉更多的是需要数学去证明。难点在于速度相同的马要不要比。
分3个步骤解这个题
先排序
（1）田最快的马大于王最快的马，直接比，田赢一场，进行下一场
（2）田最快的马小于王最快的马，将田最慢的马和王最快的马比，王赢一场，进行下一场
（3）田最快的马等于王最快的马，将田最慢的马和王最慢的马比，如果田最慢的马大于王最慢的马，那比一场，田赢一场，进行下一场。 如果田最慢的马不大于王最慢的马，则将田最慢的马和王最快的马比（如果相等进行下一场，如果王的马快，则王赢一场，进行下一场）。

大概思路就是这样，个人感觉跟需要数学推导才能出来，反正是挺不容易想到的。。。。

#include <iostream>#include <cstring>#include <algorithm>using namespace std;bool cmp(int a,int b){    return a>b;} int a[1005],b[1005];int visiteda[1005],visitedb[1005];int main(int argc, char *argv[]){    int n,sum,xx;    while(cin>>n&&n)    {        int uu=n;         xx=0;        sum=0;        memset(visiteda,0,sizeof(visiteda));        memset(visitedb,0,sizeof(visitedb));        memset(a,0,sizeof(a));              memset(b,0,sizeof(b));        for(int i=0;i<n;++i)        {            cin>>a[i];        }        for(int i=0;i<n;++i)        {            cin>>b[i];        }        sort(a,a+n,cmp);        sort(b,b+n,cmp);        int j=0,m=n;        //i表示田当前最快的马，j表示王当前最快的马,n-1表示田最慢的马，m-1表示王最慢的马         for(int i=0;i<n;++i)//sum表示田赢的场次，xx表示田输的场次         {            if(a[i]>b[j])   //田最快的马大于王最快的马             {                ++j;                 ++sum;                continue;               }            else if(a[i]<b[j])  //田最快的马小于王最快的马，将田最慢的马和王比             {                --n;                --i;                j++;                xx++;            }            else if(a[i]==b[j]) //田最快的马等于王最快的马             {                if(a[n-1]>b[m-1])//田最慢的马大于王最慢的马                 {                    --n;                    --m;                    --i;                    ++sum;                }                else if(a[n-1]==b[j])//田最慢的马等于王最快的马                 {                    --i;                    --n;                    ++j;                }                else//田最慢的马小于王最快的马                 {                    --i;                    --n;                    ++j;                    xx++;                }             }        }        cout<<200*(sum-xx)<<endl;     }     return 0;}