c++ - 最快的算法计算数组中 3 个长度 AP 的数量

Question

我要解决this CodeChef 挑战:

假设我们有一个包含 N(范围为 100,000)个元素的数组 A。我们要找到所有 3 个这样的元素对的计数 1<=Ai,Aj,Ak<=30,000 使得

Aj-Ai = Ak- Aj and i < j < k

换句话说，Ai、Aj、Ak 处于等差级数。例如数组:

9 4 2 3 6 10 3 3 10

所以 AP 是:

{2,6,10},{9,6,3},{9,6,3},{3,3,3},{2,6,10} 

所以要求的答案是 5。

我的方法

我尝试的是取名为 past 和 right 的 30,000 个长数组。最初右边包含每个 1-30,000 元素的计数。

如果我们在第i个位置，past存储i之前数组值的计数，right存储i之后数组的计数。我只是循环查找数组中所有可能的公差。这是代码:

right[arr[1]]--;

for(i=2;i<=n-1;i++)
{
    past[arr[i-1]]++;
    right[arr[i]]--;
    k=30000 - arr[i];
    if(arr[i] <= 15000)
        k=arr[i];
    for(d=1;d<=k;d++)
    {
        ans+= right[arr[i] + d]*past[arr[i]-d] + past[arr[i] + d]*right[arr[i]-d];
    }
    ans+=past[arr[i]]*right[arr[i]];
}

但这让我超出了时间限制。请提供更好的算法。

Answer 1

如果您第一次遍历列表并仅提取可能有 3 项 AP 的数字对(差异为 0 mod 2)，则可以大大缩短执行时间。然后在这些对之间迭代。

伪 C++-y 代码:

// Contains information about each beginning point
struct BeginNode {
  int value;
  size_t offset;
  SortedList<EndNode> ends;  //sorted by EndNode.value
};

// Contains information about each class of end point
struct EndNode {
  int value;
  List<size_t> offsets; // will be sorted without effort due to how we collect offsets
};

struct Result {
  size_t begin;
  size_t middle;
  size_t end;
};

SortedList<BeginNode> nodeList;
foreach (auto i : baseList) {
  BeginNode begin;
  node.value = i;
  node.offset = i's offset; //you'll need to use old school for (i=0;etc;i++) with this
  // baseList is the list between begin and end-2 (inclusive)
  foreach (auto j : restList) { 
    // restList is the list between iterator i+2 and end (inclusive)
    // we do not need to consider i+1, because not enough space for AP
    if ((i-j)%2 == 0) { //if it's possible to have a 3 term AP between these two nodes
      size_t listOffset = binarySearch(begin.ends);
      if (listOffset is valid) {
        begin.ends[listOffset].offsets.push_back(offsets);
      } else {
        EndNode end;
        end.value = j;
        end.offsets.push_back(j's offset);
        begin.ends.sorted_insert(end);
      }
    }
  }
  if (begin has shit in it) {
    nodeList.sorted_insert(begin);
  }
}
// Collection done, now iterate over collection

List<Result> res;
foreach (auto node : nodeList) {
  foreach (auto endNode : node.ends) {
    foreach (value : sublist from node.offset until endNode.offsets.last()) {
      if (value == average(node.value, endNode.value)) {
        // binary_search here to determine how many offsets in "endNode.offsets" "value's offset" is less than.
        do this that many times:
          res.push_back({node.value, value, endNode.value});
      }
    }
  }
}

return res;