Odd-Even Subsequence 解題報告

題目鏈接：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 $min(max(s1,s3,s5,…),max(s2,s4,s6,…))$.

For example, the cost of $\left\{7,5,6\right\}$ is $min(max(7,6),max(5))=min(7,5)=5$.

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

比較明顯能看得出思路是二分答案，然後一種思路是將數列排序後取若干相間隔的數，但這樣的話會有個問題：數列兩端點的數取與不取的情況很難討論。

因此考慮另一種思路：順序取數，用一個bool變量記錄當前取的是奇數位還是偶數位，對於奇數位/偶數位取數時保證其滿足小於等於mid，對於偶數位/奇數位則無限制條件。這種貪心思路爲什麼成立呢？

假設我們遇到了一個滿足條件的數，它的下一個數也滿足條件，那這時應該取哪個數呢？這個就是貪心的核心思想，顯然，取的數越前越好，因此直接取當前數即可。

代碼：

#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;
}