poj -- 2287 Tian Ji -- The Horse Racing(田忌赛马，递归分治)

Tian Ji -- The Horse RacingTime Limit: 5000MS Memory Limit: 65536K
Total Submissions: 7104 Accepted: 2178


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

 

 

题意很好理解不多说。

 

解法：1.将2组马排好序,这边我是2组都从小到大排

         2.设定2个指针s和e,表示国王那组马当前 剩下的马中最快与最慢，因此一开始s=n,e=1;

         3.从慢到快，对田忌的每匹马来跟国王的最快或最慢的马做匹配。这里有3种情况

             1)当前田忌马比国王的最慢的还慢，则败局+1，并将此马与国王剩余中最快的马比较，因为这样代价最小。同时s指向次快的马。

             2)当前田忌马比国王的最慢快,  则胜局+1，e指向次慢的马，这样保留的实力最大。

             3)当前田忌和国王的最慢马同速，利用递归进行分治。分别让当前马与最快的马比（输），或让当前马与最慢马打平，比较2者的大                 小。

 

注:必须保存状态。即田忌第A匹马与国王第B匹马比的最优解

 

CODE

 

 //4168K 125MS #include <iostream> #include <algorithm> using namespace std; #define max(a,b) a>b? a:b int a[1010]; int b[1010]; int map[1010][1010]; int n; int solve(int s,int e,int now) { if(now>n) return 0; int ans=0; if(a[now]>b[e]) { ans++,e++; if(map[s][e]==-1) map[s][e]=solve(s,e,now+1); ans+=map[s][e]; } else if(a[now]<b[e]) { ans--,s--; if(map[s][e]==-1) map[s][e]=solve(s,e,now+1); ans+=map[s][e]; } else { if(map[s-1][e]==-1) map[s-1][e]=solve(s-1,e,now+1); if(map[s][e+1]==-1) map[s][e+1]=solve(s,e+1,now+1); ans=max(map[s-1][e]+ans-1,map[s][e+1]+ans); } return ans; } int main() { int i,s,e; while (scanf("%d",&n)&&n) { s=n,e=1; for (i=1;i<=n;i++) scanf("%d",&a[i]); for (i=1;i<=n;i++) scanf("%d",&b[i]); sort(a+1,a+n+1); sort(b+1,b+n+1); memset(map,255,sizeof(map)); printf("%d/n",solve(s,e,1)*200); } }