Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11572    Accepted Submission(s): 3239

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  

 

Source 2004 Asia Regional Shanghai  

 

Recommend JGShining     貪心策略. 很容易被題目意思誤導過去用最大權值匹配。 一、如果a的最慢速度大於b的最慢，則直接a的最慢與b的最慢比賽，贏一場；
二、如果a的最慢速度小於b的最慢，則用a的最慢浪費b的最快，輸一場；
三、如果a的最慢速度等於b的最慢，則：
1.如果a的最快速度大於b的最快，則直接a的最快與b的最快進行比賽，贏一場；
2.如果a的最快速度小於b的最快，則用a的最慢浪費b的最快，輸一場；
3.如果a的最快速度等於b的最快，即a與b的最慢與最快分別相等，則：
a.如果a的最慢速度小於b的最快，則用a的最慢浪費b的最快，輸一場；
b.如果a的最慢速度等於b的最快，即a的最慢、a的最快、b的最慢、b的最快相等，
說明剩餘未比賽的馬速度全部相等，直接結束比賽。   貪心策略也容易理解。但是證明比較麻煩，不去理解了。

/*
HDU 1052
一、如果a的最慢速度大於b的最慢，則直接a的最慢與b的最慢比賽，贏一場；
二、如果a的最慢速度小於b的最慢，則用a的最慢浪費b的最快，輸一場；
三、如果a的最慢速度等於b的最慢，則：
1.如果a的最快速度大於b的最快，則直接a的最快與b的最快進行比賽，贏一場；
2.如果a的最快速度小於b的最快，則用a的最慢浪費b的最快，輸一場；
3.如果a的最快速度等於b的最快，即a與b的最慢與最快分別相等，則：
a.如果a的最慢速度小於b的最快，則用a的最慢浪費b的最快，輸一場；
b.如果a的最慢速度等於b的最快，即a的最慢、a的最快、b的最慢、b的最快相等，
說明剩餘未比賽的馬速度全部相等，直接結束比賽。

*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAXN=1010;
int a[MAXN],b[MAXN];
int main()
{
int n;
while(scanf("%d",&n)==1&&n)
{
for(int i=0;i<n;i++)scanf("%d",&a[i]);
for(int i=0;i<n;i++)scanf("%d",&b[i]);
sort(a,a+n);
sort(b,b+n);
int al=0,ah=n-1;
int bl=0,bh=n-1;
int ans=0;
while(al<=ah&&bl<=bh)
{
if(a[al]>b[bl])
{
ans+=200;
al++;bl++;
}
else if(a[al]<b[bl])
{
ans-=200;
al++;bh--;
}
else
{
if(a[ah]>b[bh])
{
ans+=200;
ah--;bh--;
}
else if(a[ah]<b[bh])
{
ans-=200;
al++;bh--;
}
else
{
if(a[al]<b[bh])
{
ans-=200;
al++;bh--;
}
else if(a[al]==b[bh])//所有的都一樣了
{
break;
}
}
}
}
printf("%d\n",ans);
}
return 0;
}