到七夕啦！在这样的日子里，Godv当然要去陪女朋友玩耍啦~~~（不要问我敬老师的手环送没送出去！）于是，资产阶级的Godv选择了带女朋友去玩赛马~~~
话说，两个人各有n匹马，每匹马各有各的速度，两个人一共比n场，且每匹马恰参加一次比赛。两个人事先约定好，对于每场比赛，获胜的得一分，平局或失败不得分，最后输掉比赛的要请对方晚上看电影。
Godv这样既聪明又萌萌哒又体贴女朋友的人，当然是不想让女朋友输啦。于是他事先要对比赛做了手脚，这样每一场比赛的参赛的马儿都是被Godv钦定好了的。
下面，你能帮Godv算一算，Godv最多能让女朋友赢多少分吗？
多组数据，数据组数不超过10组，请处理到文件结束。每组数据第一行包含一个数n（1<=n<=1e5），第二行包含n个正整数 ai (1<=ai<=1e9)，表示Godv的第i匹马的速度，第三行包含n个正整数 bi (1<=bi<=1e9)，表示Godv的女朋友的第i匹马的速度。
对于每组数据都输出一行。如果Godv的女朋友只能得到0分，请输出”Godv too strong”，否则，输出Godv的女朋友最多能得到的分数。
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 const int N=1e5+7;
5 long long x[N];
6 long long y[N];
7 int n;
8 bool mycmp (int a,int b) {
9 return (a>b);
10 }
11 int main ()
12 {
13 while (scanf("%d",&n)!=EOF) {
14 for (int i=1;i<=n;i++)
15 scanf("%lld",&x[i]);
16 for (int i=1;i<=n;i++)
17 scanf("%lld",&y[i]);
18 sort(x+1,x+1+n,mycmp);
19 sort(y+1,y+1+n,mycmp);
20 int l1,l2;
21 int h1,h2;
22 h1=h2=1;
23 l1=l2=n;
24 int ans=0;
25 while (n--) {
26 if (y[l2]>x[l1]) {
27 ans++;
28 l1--;
29 l2--;
30 }
31 else {
32 l2--;
33 h1++;
34 }
35 }
36 if (ans==0) printf("Godv too strong\n");
37 else printf("%d\n",ans);
38 }
39 return 0;
40 }
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.
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.
For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.
3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0
喜欢中文的同学，这里来。题目相似，只不过这里赢了得一分，平了不得分，输了扣一分。
可是这个扣一分，将问题又上升一个档次。
首先若我的最快马大于你最快的马，我就用我最快的马消耗你最快的马，这一点没问题。相同如果赢不了对方最快的马我就用我最慢的消耗它行吗，这个是不行的。因为我们要尽量保持失败扣分。前一个是不扣分所以没关系。举个简单的例子 2匹马（5 4） （5 3）采用一种策略 （4-5 5-3）不得分 ，实际上我们可以得到（4-3，5-5）一分。只是我们的策略是尽可能多的赢，尽可能的平，实在不行就输。策略如下：
1 若我最快的马大于对方最快的马，则消耗它。
2。1不成立下，若我最慢的马大于你最慢的马，所以我的所有马都能赢对方最慢的马，那我就用我 最慢的马消耗你。
3 1不成立下。若我最慢的马小于你最慢的马，反正我这匹马都要输就输给你最快的那一匹；
4 最难的，1不成立下，两匹最慢的马速度相同 ，分两种情况，
如对方最快的马大于我最快的马，反正都输我用我最慢的马消耗你最快的马
若两匹快马也相同， （6 43）（6 5 3）若我最慢的马采用平局（3-3）（4-6）（6-5）得分只能是0；
若用最慢的消耗对方最快的（3-6） （4-3） （6-5） 可以得一分；
意思就是说我最慢的小于你最快的，同样你之中肯定有一个小于我最快的，也许是比最慢的快一点，这样一胜一平，但结果是我吧对方 最慢的留了下来。这样更好嘛（也许有些混乱，我太菜了，欢迎交流-821474143）。
献上代码
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1007;
int a[N];
int b[N];
int n;
int main ()
{
while (~scanf("%d",&n)&&n!=0) {
for (int i=1;i<=n;i++)
scanf ("%d",&a[i]);
for (int i=1;i<=n;i++)
scanf ("%d",&b[i]);
sort (a+1,a+1+n);
sort (b+1,b+1+n);
int ans=0;
int h1,l1;
int h2,l2;
h2=h1=n;
l2=l1=1;
while (n--) {
if (a[h2]>b[h1]) {
ans++;
h2--;
h1--;
}
else if(a[l2]>b[l1]) {
ans++;
l2++;
l1++;
}
else {
if (a[l2]<b[h1]) ans--;
l2++;
h1--;
}
}
printf("%d\n",ans*200);
}
return 0;
}
小时候以为田忌赛马是一件太过简单的东西，此处向孙膑再次献上我的orz...
下期专题递归，欢迎留言哦！！！！！！！！！！！！！！！
