Monday, November 30, 2015

Buy one get one free - Zenefits


http://yuancrackcode.com/2015/10/23/buy-one-get-one-free/
一堆商品,买一个可以送一个,但送的那个的价格必须小于买的那个的价格(强调一下,不能等于)。给定商品总数n和每个商品的价格,求得到全部商品的最少开销。 例如:4个商品价格为[5, 4, 3, 3],最优解为9,即买5和4,送3和3。 
两个test case: [100, 99, 98, 1, 1, 1], [100, 99, 98, 98, 97, 97, 97, 97]
http://www.jiuzhang.com/qa/221/

这个题确实是很难的动态规划。属于区间型动态规划。在九章算法强化班中会讲到这个类型的动态规划。
首先排序不用说了。从小到大(从大到小也可以,我比较喜欢从小到大)。然后先给出DP的方程:
定义状态 f[i][j] 为 i 到 j 这一段的最优值(最少花多少钱全买下来)
    f[i][j] = min(f[i + 1][j - 1]+A[j], f[i][k] + f[k+1][j]) 其中k为i到j-1。
然后我们来证明为什么这个方程是对的。这个方程的假设前提是,不存在相交的搭配,也就是说不存在下面这种情况的匹配:
从小到大的4个数:a<= b<= c<= d。 a与c一起买,b与d一起买。
为毛呢?首先a和d肯定不相等,否则的话,abcd都相等了,这种情况是不会出现的(我们就是要证明这种情况不会出现),然后看b和c,如果他们相等,那么换成(a,b)和(c,d),代价一样,并且不会产生相交的匹配(什么叫做相交?你在ac连一条曲线,和bd之间的曲线一定相交)。然后看b和c不等的情况,不等的话,换成(b,c) + (a,d),也还是不影响结果=c+d, 然后又能使得连线不相交。
好了,上面我们证明了不相交理论。那么既然连线不想交的话,i-j这一段的匹配情况,只存在两种可能性:
  • 第一种可能是i和j匹配,然后中间的自己配对。
  • 第二种可能是,一定存在一条分割线,使得这个i-j的区间被分为i-k和k+1-j,两个区间之间的自己匹配,没有互相的连边。
好了,综上所述。问题已经解决并证明。我知道理解起来有点难,特别是证明,我也证明了好久……
顺便说一下时间复杂度 : O(n^3),可能可以用决策单调的优化优化到O(n^2)。
你可能要问我,我TMD是怎么想到这个解法的呢?因为课上讲过的DP就那么几种大类型……面试99%都落在 “划分/区间/矩阵/双序列/背包" 这五种类型。一个个试试看就行了……首先想到的是划分,然后发现推导不出来。然后矩阵肯定不是,双序列和背包也不是,只剩下区间了……
http://www.1point3acres.com/bbs/thread-143891-1-1.html
http://www.1point3acres.com/bbs/thread-143891-2-1.html
  1. int LeastCost(vector<int>& costs) {
  2.         int n = costs.size();. 1point 3acres 璁哄潧
  3.         if (n == 0)
  4.                 return 0;
  5.         sort(costs.begin(), costs.end());
  6.         vector<vector<int> > T(n, vector<int>(n, 0));. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷

  7.         for (int i = 0; i < n; i++)
  8.                 T[i][i] = costs[i];
  9.         for (int len = 2; len <= n; len++)
  10.                 for (int i = 0, j = i + len - 1; j < n; i++, j++) {. 鍥磋鎴戜滑@1point 3 acres
  11.                         T[i][j] = T[i][j-1] + costs[j];
  12.                         for (int k = i; k < j; k++) {
  13.                                 int p1 = costs[j], p2 = 0, p3 = 0;
  14.                                 if (costs[k] == costs[j])
  15.                                         p1 += costs[k];
  16.                                 if (k - 1 >= i)
  17.                                         p2 = T[i][k-1];
  18.                                 if (j - 1 >= k + 1). 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  19.                                         p3 = T[k+1][j-1];
  20.                                 if (p1 + p2 + p3 < T[i][j])
  21.                                         T[i][j] = p1 + p2 + p3;
  22.                         }
  23.                 }
  24.         return T[0][n - 1];
  25. }



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