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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. |
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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.
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Output
For each input case, output a line containing a single number,
which is the maximum money Tian Ji will get, in silver dollars.
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Sample Input
3 92 83 71 95 87 74 2 20 20 20 20 2 20 19 22 18 0 |
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Sample Output
200 0 0 |
算法分析： 1 当Tian的最快的比king的最快的马快时，让tian的最快的马和king的最快的马比；
2当tian的最快的比king的最快的马慢时，让tian的最慢的马和king的最快的马比；
3当tian的最快的马比king的最快的马相等时，再分情况比较：
1当tian的最慢的马比king的最慢的马快时，让tian的最慢的马和king的最慢的马比；
2当tian的最慢的马比king的最慢的马慢时，让tian的最慢的马和king的最快的马比；
3当tian的最慢的马和king的最慢的马相等时，让tian的最慢的马和king的最慢的马比，这时，又要分两种情况讨论：
1tian的最慢的马和king的最快的马相等
2tian的最慢的马比king的最快的马慢；
证明：
1 当Tian的最快的比king的最快的马快时，让tian的最快的马和king的最快的马比；
设tian的最快的马是t1，king的最快的马是k1，若t1不和k1比，而是和king的另一匹马k2比，k1与tian的另一匹马t2比，这时若t2》k1，和原来一样都是胜两场，若t2《k2，和原来一样都是胜一负一场，若k2《t2《k1，原来是胜两场现在是胜一负一. 得证,
2当tian的最快的比king的最快的马慢时，让tian的最慢的马和king的最快的马比；
设tian的最快的马是t1，king的最快的马是k1，tian的最慢的马是t2。第一感觉是反正与king的最快的马比，怎么比都输，那么就用代价最少的，让tian的最慢的马和king的最快的马比。然后一想，上面的证明同样可以证明当tian的最慢的马比king的最慢的马快时，则让tian的最慢的马和king的最慢的马比，这样好想第一种想法有点漏洞，然后仔细分析，若tian的最慢的马比king的最慢的马快，则不管怎么比，king的最慢的马都输，那么就无所谓了。
3当tian的最快的马比king的最快的马相等时，这时并不能确定一种比较的方案，所以再分情况比较：
1当tian的最慢的马比king的最慢的马快时，让tian的最慢的马和king的最慢的马比；这个的证明和上面第一个证明原理一样。
2当tian的最慢的马比king的最慢的马慢时，让tian的最慢的马和king的最快的马比；这个的证明和上面第二个证明原理一样。
3当tian的最慢的马和king的最慢的马相等时，让tian的最慢的马和king的最慢的马比
当用最慢的和最慢的比，最快的和最快的比时，是一个平局。而用tian的最慢的和king的最快的比时，这时tian剩下所有的马都比king的最慢的马快，所以这时也是一个平局，但是这时tian的最快的马保留了下来，所以得证。
这种题好像暗含着一种处世的道理，有时，尤其是当我们无法看到很远处是什么样子的时候，我们只能一步一步的做对的事情，这时会遇到想上面最快的马和最快的马速度相等的处境，这时，我们不能因为急着找下一步该干什么而急的团团转，这时也许去其它方面想想办法，就又知道该怎么做了。
#include<stdio.h>
#include<stdlib.h>
void main()
{
int *t,*k;
int n;
int end,tl,th,kl,kh;
while(1)
{
scanf("%d",&n);
if(n==0)
return;
t=(int *)malloc(n*sizeof(int));
k=(int *)malloc(n*sizeof(int));
for(int i=0;i<n;i++)
scanf("%d",&t[i]);
for(i=0;i<n;i++)
scanf("%d",&k[i]);
for(i=0;i<n-1;i++)
{
int temp1=t[i],temp2=k[i];
int x1,x2;
for(int j=i+1;j<n;j++)
{
if(t[j]>temp1)
{
temp1=t[j];
x1=j;
}
if(k[j]>temp2)
{
temp2=k[j];
x2=j;
}
}
if(t[i]!=temp1)
{
int temp=t[x1];
t[x1]=t[i];
t[i]=temp;
}
if(k[i]!=temp2)
{
int temp=k[x2];
k[x2]=k[i];
k[i]=temp;
}
}
tl=kl=0;
th=kh=n-1;
end=0;
for(i=0;i<n;i++)
{
if(t[tl]>k[kl])
{
tl++;
kl++;
end+=200;
}
else if(t[tl]<k[kl])
{
kl++;
th--;
end-=200;
}
else
{
if(t[th]>k[kh])
{
th--;
kh--;
end+=200;
}
else if(t[th]<k[kh])
{
th--;
kl++;
end-=200;
}
else
{
if(t[th]==k[kl])
{
th--;
kl++;
}
else if(t[th]<k[kl])
{
th--;
kl++;
end-=200;
}
}
}
}
printf("%d
",end);
}
}