Tian Ji -- The Horse Racing

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 19   Accepted Submission(s) : 5

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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

题意:
田忌赛马的故事大家都知道吧,现在你要做的工作就是,计算田忌能赚多少钱,赢一场200,平一场0,负一场-200.
分析:
这是一道贪心算法题。
要用田忌快的马去跟国王的马比。
一步一步实现。
注意:
Solution:这题有多种解体思路,DP,二分图最大匹配算法等,这里给出的是比较容易理解的贪心算法
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int t[3001],k[3001];
int main(){
    int n,i,j,a,b,w;
    while(scanf("%d",&n),n){
        for(i=0;i<n;i++)
            scanf("%d",&t[i]);
        for(i=0;i<n;i++)
            scanf("%d",&k[i]);
            w=0;
        sort(t,t+n);
        sort(k,k+n);//sort的默认排列顺序是升序。
        for(i=0,j=n-1,a=0,b=n-1;a<=b&&i<=j;)//从两头逼近,两边同时缩小范围
            {                 if(t[i]>k[a])//最慢的跟最慢的比,如果快就sum++
                {
                    i++;
                    a++;
                    sum++;
                }
                else if(t[i]<k[a])//再比倒数第二组,如果田忌的马还是慢的话,
                {
                    i++;
                    b--;
                    sum--;
                }
                else if(t[j]>k[b])//快的跟快的比
                {
                    j--;
                    b--;
                    sum++;
                }
                else if(t[j]<k[b])
                {
                   i++;
                   b--;
                   sum--;
                }
                else if(t[i]<k[b])
                {
                    i++;
                    b--;
                    sum--;
                }
                else
                {
                    i++;
                    b--;
                }
            }
            printf("%d\n",sum*200);
    }
    return 0;
}
未完待续,DP,二分图最大匹配算法。

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