HDU 1052 Tian Ji -- The Horse Racing(贪心)

Tian Ji – The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31951    Accepted Submission(s): 9696

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

 

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题意

    田忌赛马。田忌和齐王各有n匹马，输入田忌的马的速度和齐王的马的速度。每一轮田忌赢了就得200两银子，平就得0两，输了就失去200两银子。问田忌最多能得到多少。

思路

    这题可以用很多种方法做，我是用贪心的。

1、如果田忌目前还未赛跑过的最快的马比齐王最快的马快，那么就先赢一场；
2、如果田忌目前还未赛跑过的最慢的马比齐王最慢的马快，那么也可以赢一场；
3、如果田忌目前还未赛跑过的最慢的马比齐王最慢的马慢，那么把它拿出来和齐王最快的马比，输一场；
4、如果田忌目前还未赛跑过的最快的马比齐王最快的马慢，那么拿最慢的马和齐王最快的马比，输一场；
5、如果田忌目前还未赛跑过的最快的马和齐王最快的马相等时，那么拿最慢的马来和齐王最快的马比。

正确性么..我就不证明了，相比大家推一下都知道的（逃。

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
int n,a[1005],b[1005];
int main() {
    while(1) {
        readInt(n);
        if(!n)
            break;
        for(int i=1;i<=n;i++)
            readInt(a[i]);
        for(int i=1;i<=n;i++)
            readInt(b[i]);
        sort(a+1,a+n+1);
        sort(b+1,b+n+1);
        int num=0,mxt=n,mxk=n,mit=1,mik=1;
        while(mit<=mxt) {
//          cout<<mit<<" "<<mxt<<" "<<mik<<" "<<mxk<<" "<<a[mxt]<<" "<<b[mxk]<<endl;
            if(a[mxt]>b[mxk]) {
                num++;
                mxk--;
                mxt--;
                continue;
            }
            if(a[mxt]<b[mxk]) {
                num--;
                mit++;
                mxk--;
                continue;
            }
            if(a[mit]>b[mik]) {
                num++;
                mit++;
                mik++;
                continue;
            }
            if(a[mit]<b[mxk])
                num--;
            mit++;
            mxk--;
        }
        printf("%d\n",200*num);
    }
}