hdu 1052 Tian Ji -- The Horse Racing（贪心算法）

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Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30233    Accepted Submission(s): 9117

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

 

Source

2004 Asia Regional Shanghai

 

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本题题意：很有名的历史性问题，相信大家不用看题目，光看标题和图片就能知道本题的大致题意，那就是著名的田忌赛马问题。田忌和他的国王赛马，田忌每赢一场就增加200美金，规定每匹马都只能出场一次，求田忌能拿到的最好结果，即最优解。

解题思路：很容易看出这是一个贪心的算法，只不过这里需要考虑的情况比较的多，具体方案请看代码中的注解，在此之前我们需要提前将他们两人的马匹排下序，这里我用了手写快排的方法，因为我那时刚好学了一下，本题是我以前做过的题目，当然大家也可以直接调用sort函数就好，但需要注意的是比较顺序，我前面用的是从大到小的顺序，那么调用sort()快排函数也一样要从大到小排序，这就需要再写一个cmp的函数，因为sort（）函数默认是从小到大排序，若大家直接用sort()函数，那么最开始应该从数组尾部开始比较。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>           //sort函数头文件using namespace std;bool cmp (const int a, const int b){    return a > b;}void qk(int a[],int l,int r)  //手写快排函数{    int i,j,t;    if(l<r)    {        i=l;        j=r;        t=a[i];        while(i<j)        {            while(i<j&&a[j]<=t)                j--;            a[i]=a[j];            while(i<j&&a[i]>t)                i++;            a[j]=a[i];        }        a[i]=t;        qk(a,l,i-1);        qk(a,i+1,r);    }}int main(){    int n,a[1005]= {0},b[1005]= {0};    while(~scanf("%d",&n)&&n)    {        int i,j,la,lb,tj=0,King=0,c=0;        for(i=0; i<n; i++)            scanf("%d",&a[i]);        for(j=0; j<n; j++)            scanf("%d",&b[j]);        qk(a,0,n-1);  //从大到小排序，手写快排        qk(b,0,n-1);       //sort(a,a+n,cmp);       //sort(b,b+n,cmp);        i=j=0;        la=lb=n-1;        while(i!=la)        {            if(a[i]>b[j])   //田忌最快的马比齐王最快的马快            {                tj++;        //直接赢，往后再比较                i++;                j++;            }            else if(a[la]>b[lb]) //否则比较田忌与齐王最慢的马，田忌最慢的马比齐王慢快的马快            {                tj++;            //也直接赢，往前比较                la--;                lb--;            }            else if(a[la]<b[j]) //否则就拿田忌此时最慢的马与齐王最快的马慢（此时田忌最快的马比齐王最快的马慢）            {                King++;            //直接输                la--;                j++;            }            else                break;        }        if(a[i]>b[j])           //剩下最后一匹马再比一下就可以了（因为上题条件是i=la时退出循环）            tj++;        else if(a[i]<b[j])            King++;        c=200*(tj-King);        printf("%d\n",c);    }    return 0;}