hdu 1052 Tian Ji -- The Horse Racing

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

392 83 7195 87 74220 2020 20220 1922 180

 

Sample Output

20000

 

这个题真的很棒，虽然自己没想出来，比较菜，看了大神键盘上的舞者的csdn博客，自己才想懂，一开始自己的思路太狭隘，得向人家学习，解题得完整，符合任何一种情况才是真的解决方案

首先用l1跟r1来表示田忌的情况，l2跟r2来表示国王的情况，分析

如果田忌最快的马跟国王最快的马比较，田忌赢得话，就比较后面的;

如果输的话，就让国王最快的马跟田忌最慢的马比较输一场，因为反正都得输，如果用最慢的马输了，其他的马也许还能比国王的慢马快;

如果快马都一样，那就让慢马比较，又分三种情况，如果田忌的慢马比国王的快，那就赢一局；

如果慢的话，那就代表着必输的状态了，那就让田忌的慢马去跟国王的快马比较呗，输一局；

如果慢马跟快马都一样的话，那就再次判断田忌的慢马跟国王的快马的情况，或者就让他们去平一局，

总结：能赢就尽量赢，必输的话就让田忌的慢马去跟国王的快马比赛，实在不行就平局。

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>#include<math.h>using namespace std;#define ll long longconst int INF=0x7fffffff;const int N=1001;int a[N],b[N];int main(){    int n;   while (~scanf("%d",&n),n)    {        int i,j,sum=0;        int l1,l2,r1,r2;        for (i=0;i<n;i++)            scanf("%d",&a[i]);        for (i=0;i<n;i++)            scanf("%d",&b[i]);            sort(a,a+n);            sort(b,b+n);        l1=l2=0;        r1=r2=n-1;      while (l1<=r1&&l2<=r2)      {          if (a[r1]>b[r2])          {              sum++;              r1--;              r2--;          }else if (a[r1]<b[r2])            {                sum--;                l1++;                r2--;            }          else{             if (a[l1]>b[l2])             {                sum++;                l1++;                l2++;             }else if (a[l1]<b[l2])             {                sum--;                l1++;                r2--;             }             else{                if (a[l1]<b[r2])                    sum--;                 l1++;                 r2--;             }            }      }      printf("%d\n",sum*200);    }    return 0;}