Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37056    Accepted Submission(s): 11153

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     題目大意： 田忌和齊威王賽馬。兩人各有n匹馬。每匹馬都有一個速度屬性。賽馬時如果平局則不計分，如果贏了記200分，如果輸了扣200分。問怎樣匹配田忌分最高，輸出最高分。   貪心好題。 首先比較田忌和齊威王最慢的馬，分別為T和Q。 如果T快，則讓T贏Q。因為田忌的馬都能贏Q，所以用最少的耗費贏Q； 如果T慢，則讓T輸給齊威王最快的馬。因為齊威王的馬都能贏T，所以拉齊威王最快的馬下水； 如果T和Q一樣快，則比較田忌和齊威王最快的馬，分別為TT和QQ： 如果TT快，則讓TT贏QQ，因為贏誰都是贏，所以贏齊威王最快的馬； 如果TT慢，則讓QQ贏田忌最慢的馬，因為誰輸QQ都是輸，用最小的耗費輸給他； 如果TT和QQ一樣快，則讓T輸給QQ，因為讓T、TT平Q、QQ，則田忌並不賺，用T輸給QQ，再用一個較小的數贏Q，則田忌省下了TT，穩賺。  

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>

typedef long long lol;

using namespace std;

const int maxn=1000;

int a[maxn+5];
int b[maxn+5];

int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=n;i++)
            scanf("%d",b+i);
        sort(a+1,a+1+n);
        sort(b+1,b+1+n);

        int win=0,tie=0,los=0;
        int a1=1,b1=1,a2=n,b2=n;
        while(a1<=a2)
        {
            if(a[a1]>b[b1])
            {
                win++;
                a1++;b1++;
            }
            else if(a[a1]<b[b1])
            {
                los++;
                a1++;b2--;
            }
            else
            {
                if(a[a2]>b[b2])
                {
                    win++;
                    a2--;b2--;
                }
                else if(a[a2]<b[b2])
                {
                    los++;
                    a1++;b2--;
                }
                else
                {
                    if(a[a1]<b[b2])
                    {
                        los++;
                        a1++;b2--;
                    }
                    else
                    {
                        tie++;
                        a1++;b2--;
                    }
                }
            }
        }

        printf("%d\n",win*200-los*200);
    }
    return 0;
}

View Code