poj 2287（贪心）

　Tian Ji -- The Horse Racing

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 12490		Accepted: 3858

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

Shanghai 2004

 

思路：现将两人马按速递增排列；

1）若田忌最慢的马快于齐王的最慢的马（a1>b1),将a1与b1比，因为齐王最慢的马b1一定输，输给田忌最慢的马最好；

2）若田忌最慢的马慢于齐王的最慢的马（a1<b1),将a1与bn比，因为田忌最慢的马a1一定输，输给齐王最快的马最好；

3）若田忌最快的马快于齐王的最快的马（an>bn),将an与bn比，因为田忌最快的马an一定赢，赢齐王最快的马最好；

4）若田忌最快的马慢于齐王的最快的马（an<bn),将a1与bn比，因为齐王最快的马bn一定赢，赢田忌最慢的马最好；

5）当田忌最慢的马与齐王最慢的马相等（a1=b1)，且田忌最快的马比齐王最快的马快时（an>bn）,将an与bn比；相反，若（an<bn),则让a1与bn比；

6）田忌最快的马与齐王最快的马相等时（an==bn),将a1与bn比有最优解；

 

 

过程：田忌第一步的贪心选择是派出最快的或最慢的马与齐王的最慢的马比，得出的第一个子问题的最优解；第二步的贪心选择是在剩下的n-1匹马中派出最快的马或者最慢的马与齐王次慢的马比，

得出第二个子问题的最优解……每次贪心选择都将当前问题归纳为更小的相似问题，而每个贪心选择都仅做一次，所有子问题的最优解构成整个问题的最优解。

 

在poj上AC了，但在zoj上WA了。。。。有毒；

这道题还有种dp的解法稍后补上；

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<iomanip>
 #include<cmath>
 #include<vector>
 #include<queue>
 #include<stack>
 using namespace std;
 #define PI 3.141592653589792128462643383279502
 int main(){
     //#ifdef CDZSC_June
     //freopen("in.txt","r",stdin);
     //#endif
     //std::ios::sync_with_stdio(false);
     int n;
     int tian[],qi[];
     while(scanf("%d",&n),n){
         for(int i=;i<=n;i++)
             cin>>tian[i];
         for(int i=;i<=n;i++)
             cin>>qi[i];
         sort(tian+,tian+n+);
         sort(qi+,qi+n+);
         int t1=,tn=n,q1=,qn=n;
         int sum=;

         while(t1<=tn){
             if(tian[t1]>qi[q1]){
                 t1++;q1++;sum+=;
             }
             else if(tian[t1]==qi[q1]){
                 while(t1<=tn&&q1<=qn){
                     if(tian[tn]>qi[qn]){
                         tn--;qn--;sum+=;
                     }
                     else {
                         if(tian[t1]<qi[qn]) sum-=;
                         t1++;qn--;break;
                     }
                 }
             }
             else{
                 t1++;qn--;sum-=;
             }
         }
         cout<<sum<<endl;
     }
     return ;
 }