田忌赛马（编程）

如题：

田忌赛马

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

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.

输入

The input consists of many 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.

输出

For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.

样例输入

392 83 7195 87 74220 2020 20220 1922 18

样例输出

20000

题目是英文的，看似比较复杂，其实很大一部分都是在叙述古时候田忌赛马的故事，题目的大致意思就是，田忌和齐王各自有很多马（不止三匹），现在让他们的马比赛，

每人个出一匹，以此比赛，直到所有的马都比完，每局输的一方给赢的一方200金，平局什么都不给，求田忌最后最多能得到多少金。

       对于这个问题，很多人刚开始可能想到的解决方案是，先让各个队的马以此从高到低排好顺序，如：

田忌：98 95 92 85

齐王：99 97 96 89

然后把数存到数组里面，再弄一个双重循环：

tian[4]={98,95,92,85}     qi[4]={99,97,96,89}

int m=0,t=0;//m的作用是每次找到田忌的马大于齐王的时候，就让下一次循环从齐王的下一匹马开始，t是赢的局数。

for(int i=0;i<4;i++)

for(int j=m;j<4;j++)

{

if(tian[i]>qi[j])

{

t++;

j++;

m=j;

break;

}

}

仔细观察这段程序，其实就会发现，此代码只能求出田忌赢的局数，并不能求出输的局数。

千万不能理解为，只要求出赢的局数，拿总数减去赢的局数，就是输的局数，因为还有平局。

所以，该题的思路应该是分几种情况：

先让双方最差的马比：

如果田忌赢齐王 赢一局

如果田忌输齐王 让田忌最差的马和齐王最好的马比 输一局（但是消耗了齐王最好的一匹马）

如果平局的话 就先不管它 而是让双方的两匹最好的马相比

如果田忌赢齐王 赢一局

否则（在双方最差的马平局的情况下，田忌最好的马小于或等于齐王的马，用田忌最差的马和齐王最好的比）

{

如果田忌的最差的马赢齐王最好的马 赢一局

如果田忌的最差的马输 则输一局

否则 平局

}

看如下代码：

#include<stdio.h>
#include<stdlib.h>
int comp(const void *a,const void *b)
{
return *(int *)b-*(int *)a;
}
int main(void)
{
int num;
while(scanf("%d",&num)!=EOF)
{
int m[1001],n[1001];


for(int i=0;i<num;i++)
scanf("%d",&m[i]);
for(int j=0;j<num;j++)
scanf("%d",&n[j]);
qsort(m,num,sizeof(m[0]),comp);
//for( i=0;i<num;i++)
//printf("%d ",m[i]);
qsort(n,num,sizeof(n[0]),comp);
//for(j=0;j<num;j++)
//printf("%d ",n[j]);
int a=0;
int b=0;
int a1=num-1;
    int b1=num-1;
int t=0;
while(a<=a1)
{
if(m[a1]>n[b1])
{
a1--;
b1--;
t++;
}
else if(m[a1]<n[b1])
{
a1--;
b++;
t--;
}
else if(m[a]>n[b])
{
a++;
b++;
t++;
}
else
{
if(m[a1]>n[b])
{
b++;
a1--;
t++;
}
else if(m[a1]<n[b])
{
b++;
a1--;
t--;
}
else
{
b++;
a1--;
}
}
}
printf("%d\n",200*t);
}
return 0;
}