Odd-Even Subsequence 解题报告

最新推荐文章于 2021-10-22 12:37:32 发布

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本文解析了Codeforces竞赛中一道关于寻找子序列最小成本的问题，通过二分查找和贪心策略，实现了一种有效的解决方案。

题目链接：https://codeforces.ml/contest/1370/problem/D

Ashish has an array $a$ of size $n$ .

A subsequence of $a$ is defined as a sequence that can be obtained from $a$ by deleting some elements (possibly none), without changing the order of the remaining elements.Consider a subsequence $s$ of $a$ . He defines the cost of $s$ as the minimum between:

The maximum among all elements at odd indices of $s$ .
The maximum among all elements at even indices of $s$ .

Note that the index of an element is its index in $s$ , rather than its index in $a$ . The positions are numbered from $1$ . So, the cost of $s$ is equal to $m i n (m a x (s 1, s 3, s 5, \dots), m a x (s 2, s 4, s 6, \dots))$ .

For example, the cost of $\left\{7,5,6\right\}$ is $m i n (m a x (7, 6), m a x (5)) = m i n (7, 5) = 5$ .

Help him find the minimum cost of a subsequence of size $k$ .

比较明显能看得出思路是二分答案，然后一种思路是将数列排序后取若干相间隔的数，但这样的话会有个问题：数列两端点的数取与不取的情况很难讨论。

因此考虑另一种思路：顺序取数，用一个bool变量记录当前取的是奇数位还是偶数位，对于奇数位/偶数位取数时保证其满足小于等于mid，对于偶数位/奇数位则无限制条件。这种贪心思路为什么成立呢？

假设我们遇到了一个满足条件的数，它的下一个数也满足条件，那这时应该取哪个数呢？这个就是贪心的核心思想，显然，取的数越前越好，因此直接取当前数即可。

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
 struct newdata
 {
  int label;
  long long x;
 };
 newdata a[200005];
 int b[200005];
 int n,k;
bool cmp(newdata i,newdata j)
{
 return i.x < j.x;
}
int main()
{
 scanf("%d%d",&n,&k);
 for (int i = 1;i <= n;i ++)
 {
  a[i].label = i;
  scanf("%I64d",&a[i].x);
  b[i] = a[i].x;
 }
 sort(a + 1,a + n + 1,cmp);
 int l = 1;
 int r = n;
 while (l < r)
 {
  int mid = l + r >> 1;
  int cnt = 0;
  bool now = false;
  for (int i = 1;i <= n;i ++)
   if (now)
   {
    cnt ++;
    now ^= 1;
   }
   else
   {
    if (b[i] <= a[mid].x)
    {
     cnt ++;
     now ^= 1;
    }
   }
  if (cnt >= k)
  {
   r = mid;
   continue;
  }
  cnt = 0;
  now = true;
  for (int i = 1;i <= n;i ++)
   if (now)
   {
    cnt ++;
    now ^= 1;
   }
   else
   {
    if (b[i] <= a[mid].x)
    {
     cnt ++;
     now ^= 1;
    }
   }
  if (cnt >= k)
   r = mid;
  else l = mid + 1;
 }
 printf("%d",a[l].x);
 return 0;
}