hdu 1052（田忌赛马详解）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 (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

题意：田忌和国王赛马，输失去200金币，赢得到200，平局0金币；第一行n马，第二行田忌的n匹马的速度，第三行国王马；n=0表示结束；

思路：坑点就是平局的时候，坑我的两组数据

6   8    10                    6     7      10

5   7    10                    5     8      10

两人最快的马进行比较，田忌的马赢了就战胜了国王，先最快的马，输了就让田忌最慢的马消耗国王最快的马，平局就考虑最慢马的情况，这时切勿让田忌最慢的马直接输给国王最快的马，如果田忌最慢的马赢了国王最慢的马，就赢下这一局（把最快的马之间的比较，先放放），如果田忌最慢的马输了，让田最慢的马消耗国王最快的马，如果最慢的的比较还是平局，考虑是不是最后一次比较，如果不是，直接让田忌最慢的吗输给国王最快的马，如果是，就平局处理；详见代码注释；

最后总结一下贪心：局部最优，你的每一步都是最优的，不需要遍历全部找最优，从大体到局部出发，每一步都是当前最好的处理方式，局部最优；

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1010],b[1010];int main(){int n,i,j;while(scanf("%d",&n)!=EOF){if(n==0) break;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);int lose=0,win=0,t=0,tj=n-1,g=0,gw=n-1;for(;t<=tj;)//结束的条件； {if(a[tj]>b[gw])//田忌最快的马和国王最快的马比较； {win++;tj--;gw--;}if(a[tj]<b[gw])// 田忌最慢的马和国王最快的马比较；{lose++;t++;gw--;}if(a[tj]==b[gw])//最快的马相等，比较最慢的马； {if(a[t]>b[g])//先让田忌最慢马的赢一场； {t++;g++;win++;}if(a[t]<b[g])//让田忌最慢的马消耗国王最快的马； {lose++;t++;gw--;}if(a[t]==b[g])//坑点所在； {if(a[t]==b[gw])//有可能是最后一对马比较，t==tj 时； {t++;gw--;}else//t!=tj 时，直接让田忌这匹马输掉； {lose++;t++;gw--;}}}}printf("%d\n",(win-lose)*200);}return 0;}