总时间限制: 5000ms 内存限制: 65536kB
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 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.
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
3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0
样例输出
200
0
0
来源
Shanghai 2004
说实话 这个题除了题面描述有点恶心外,思路和当年课文中学到的一模一样,只要各种情况都想到了这个题就很好了
首先将自己的和敌方的馬排序
然后我方最差的和地方最差的能比过吗?
能的话看下一匹,我方+1
不能的话将它废物利用和对面最强的比,消耗,对面+1
平局的话
判断当前我方最快的马的情况,如果他比敌方的马还强,则不替换直接上(因为对面不管上哪个都赢不了),我方+1
否则就上最猜的消耗对面的,敌方+1
最后使用我方赢得数量-敌方赢得数量 *200就是答案
说明:代码不是本人写的,写好后看别人思路,太牛逼了,我当时枚举的代码比这长多了,意会咯
代码如下
awin=bwin=0;
afast=bfast=n-1;
aslow=bslow=0;
for(int i=0;i<n;++i){
if(a[aslow]>b[bslow]){
awin++;
aslow++; bslow++;
}
else if(a[aslow]<b[bslow]){
bwin++;
aslow++; bfast--;
}
else{
if(a[afast]>b[bfast]){
awin++;
afast--; bfast--;
}
else if(a[aslow]<b[bfast]){
bwin++;
aslow++; bfast--;
}
}
}