hdu-1052-Tian Ji -- The Horse Racing(经典)

/*
hdu-1052
Tian Ji -- The Horse Racing
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14456    Accepted Submission(s): 4125

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

Recommend
JGShining

题意:
田忌赛马

解题报告:
关键在于田忌最慢的马，能先赢就先赢，不能赢就去消耗齐王最快的马,
然后再来考虑最快的马，能先赢就先赢，不能赢说明现在田忌和齐王最快的
马和最慢的马都相等，再来考虑把田忌最慢的马和齐王最快的马比较。
可能更小，但也可能相等。(要按顺序哦，亲。)
1.当田忌最慢的马比齐王最慢的马快，赢一场先
2.当田忌最慢的马比齐王最慢的马慢，和齐王最快的马比，输一场
3.当田忌最快的马比齐王最快的马快时，赢一场先。
4.当田忌最快的马比齐王最快的马慢时，拿最慢的马和齐王最快的马比，输一场。
5.当田忌最快的马和齐王最快的马相等时，拿最慢的马来和齐王最快的马比.

//以下是网上的证明

证明：田忌最快的马和齐王最快的马相等时拿最慢的马来和齐王最快的马比有最优解。

1）假设他们有n匹马，看n=2的时候.

a1 a2
b1 b2

因为 田忌最快的马和齐王最快的马相等
所以a1=b1,a2=b2 所以这种情况有2种比赛方式，易得这两种方式得分相等。

2）当数列a和数列b全部相等等时(a1=b1,a2=b2...an=bn)，
显然最慢的马来和齐王最快的马比有最优解,可以赢n-1长，输1场，找不到更好的方法了。
3）当数列a和数列b元素全部相等时(a1=b1=a2=b2...=an=bn)，无法赢也不输。

现在假设n匹马时拿最慢的马来和齐王最快的马比有最优解，
证明有n+1匹马时拿最慢的马来和齐王最快的马比也有最优解。
数列
a1 a2 a3 a4...an an+1
b1 b2 b3 b4...bn bn+1

其中ai>=ai-1,bi>=bi-1

数列a和数列b不全部相等时，拿最慢的马来和齐王最快的马比数列得到数列
(a1) a2 a3 a4...an an+1
b1 b2 b3 b4...bn (bn+1)

分4种情况讨论
1.b1=b2,an=an+1
则有
a2 a3 a4...an
b2 b3 b4...bn
其中a2>=a1,a1=b1,b1=b2,得a2>=b2(此后这种大小关系不再论述）,an>=bn.
此时若a2=b1,根据归纳假设，有最优解，否则a2>根据前面“公理”论证有最优解。
当且仅当a数列,b数列元素全部相等时有an+1=b1,已证得，所以an+1>b1，
赢回最慢的马来和齐王最快的马比输的那一场。

2.b1<=b2,an=an+1
交换 b1,b2的位置，
数列
(a1) a2 a3 a4...an an+1
b2 b1 b3 b4...bn (bn+1)
此时 a2>=a1,an>=bn,
对于子表
a2 a3 a4...an
b1 b3 b4...bn
根据前面“公理”或归纳假设，有最优解。
an+1>=b2, 当且仅当b2=b3=b4=..=bn+1时有an+1=b2，这种情况，
a中其它元素<=b1,b2,b3,b4..bn,对于这部分来说，能赢 x盘(x<=n)，
假如不拿最慢的马来和齐王最快的马比则拿最快的马来和齐王最快的马比，
此时平一盘，能赢x-1盘，而拿最慢的马来和齐王最快的马 比，输一盘能赢x盘，
总的来说，还是X这个数，没有亏。

3.b1=b2,an<=an+1
4.b1<=b2,an<=an+1证明方法类似，不再重复。

以证得当有n+1匹马的时候，田忌和齐王最快最慢的马速度相等时，
拿最慢的马来和齐王最快的马比有最优解，已知当n=2时成立，
所以对于n>2且为整数（废话，马的只数当然是整数）时也成立。
当n=1时....这个似乎不用讨论.
*/
#include <iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#define maxn 2600
int a[maxn],b[maxn];
int main()
{
    int N,i,j,i1,j1,sum;
    while(scanf("%d",&N),N)
    {
        for(i=; i<N; i++)
            scanf("%d",&a[i]);
        for(i=; i<N; i++)
            scanf("%d",&b[i]);
            sort(a,a+N);
            sort(b,b+N);
            sum=;
            for(i=,j=N-,i1=,j1=N-;i1<=j1&&i<=j;)
            {

                if(a[i]>b[i1])
                {
                    i++;
                    i1++;
                    sum++;
                }
                else if(a[i]<b[i1])
                {
                    i++;
                    j1--;
                    sum--;
                }
                else if(a[j]>b[j1])
                {
                    j--;
                    j1--;
                    sum++;
                }
                else if(a[j]<b[j1])
                {
                   i++;
                   j1--;
                   sum--;
                }
                //现在剩余就是最快的和最慢的都相等的情况,
                //则把田忌最慢的和齐王最快的比较
                else if(a[i]<b[j1])
                {
                    i++;
                    j1--;
                    sum--;
                }
                else
                {
                    i++;
                    j1--;
                }
            }
            printf("%d\n",sum*);
    }
    return ;
}
/*
5
100 99 98 97 96
99  99 98 95 94
600
*/