ACM: LA 3266 -&nb…

Tian Ji -- The Horse Racing

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
integern ( n 1000)
on the first line, which is the number of horses on each side. The
nextn integers on the second line are the
speeds of Tian's horses. Then the
next nintegers 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: 经典问题田忌赛马, 如果田忌胜获得200, 平0, 输-200. 计算出田忌最多可以获得多少银子.

 

解题思路:

     
1. 一道DP题, 先找出决策, 再列动态方程.

     
2. 将2人得马先按照从大到小排序.

      (1)当田忌的当前最强马跟King的当前最强的马,
如果输的话,反正结果都是输, 那不如选择最差的

     
马去, 这样就会不浪费, 同时为后面的比赛准备.

     
(2)如果是胜利的话, 田忌当前最强的马肯定比King后面的马还要强. 这样当然选择当前最强的马去.

     
分析到这里应该可以知道决策有: 选出当前最强的或者是当前最差的马.

     
设dp[i][j]: King选出了i匹马, 那么田忌也应该出i匹马, 只是田忌从前面和后面怎么选的问题,

      田忌从前面选择了j匹马的最优值.

     
3. 因此, 当前King出的是i号马, 田忌要么选j号马, 要么是n-(i-j)+1号马.

        
dp[i][j] = max( dp[i-1][j-1]+判断(j, i), dp[i-1][j]+判断(n-(i-j)+1, i)
)

        
结果: max(dp
[i]) (i ∈ 1...n);

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define MAX 1005

int n;

int a[MAX], b[MAX];

int dp[MAX][MAX];

inline int max(int a, int b)

{

 return a > b ? a : b;

}

bool cmp(int a, int b)

{

 return a > b;

}

int judge(int i, int j)

{

 if(a[i] > b[j]) return 1;

 else if(a[i] < b[j]) return
-1;

 else return 0;

}

int main()

{

// freopen("input.txt", "r", stdin);

 int i, j;

 while(scanf("%d", &n) !=
EOF)

 {

  if(n == 0) break;

  for(i = 1; i <=
n; ++i)

   scanf("%d",
&a[i]);

  for(j = 1; j <=
n; ++j)

   scanf("%d",
&b[j]);

  sort(a+1, a+n+1, cmp);

  sort(b+1, b+n+1, cmp);

  dp[0][0] = 0;

  for(i = 1; i
<= n; ++i)

  {

   dp[i][0] =
dp[i-1][0] + judge(n-i+1, i);

   for(j = 1; j
< i; ++j)

   {

    dp[i][j]
= max(

       dp[i-1][j-1]+judge(j,
i),

       dp[i-1][j]+judge(
n-(i-j)+1, i )

      );

   }

   dp[i][i] =
dp[i-1][i-1]+judge(i,i);

  }

  int ans = dp
[0];

  for(i = 1; i <=
n; ++i)

   ans =
max(ans, dp
[i]);

  printf("%d\n",
ans*200);

 }

 return 0;

}