Codeforces 900C - Remove Extra One（思维好题）

Remove Extra One
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are given a permutation $p$ of length $n$. Remove one element from permutation to make the number of records the maximum possible.
We remind that in a sequence of numbers $a1, a2, ..., ak$ the element $ai$ is a record if for every integer $j (1 ≤ j < i)$ the following holds: $aj < ai$.

input
The first line contains the only integer $n (1 ≤ n ≤ 10^5)$ — the length of the permutation.
The second line contains $n$ integers $p1, p2, ..., pn$ $(1 ≤ pi ≤ n)$ — the permutation. All the integers are distinct.

Output
Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples
Input
1
1
Output
1
Input
5
5 1 2 3 4
Output
5

Note
In the first example the only element can be removed.

解题思路

先给一波官方题解：

In this problem you have to find an element after which removal the number of records is maximum possible.
Let ri be an array consisting of 0 and 1 depending on whether the $i-th$ element was a record initially or not. We can compute it easily in $O(N)$.
Let $xi$ be the difference between the number of records after removal the i-th element and initial number of records.
Let’s think of how does removal of ai influence the array $ri$. First of all, $ri$ becomes 0. $rj (j < i)$ do not change in this case. Some of $rj (j > i)$, change from 0 to 1. These elements are not records initially, but become records after the removal. These elements are elements which have only one greater element in front of them — $ai$.
Here follows an $O(n^2)$ solution. Let’s fix $ai$ — the element we are going to remove. Let $xi =  - ri$  +  the number of such $j$ that $j > i$, $ai > aj$, and for all $k (k ≠ i, k < j)$ $ak < aj.$ We can compute this just looping through all $j$ and keeping the maximum over all elements but the $i-th$.
Now note that it’s not required to fix $ai$. $rj$ can become 1 from 0 only when a certain element from the left is removed. Let’s loop through all $aj$ and determine if there is an element to the left such that it is greater than $aj$, but all other elements are less than $aj$. We can check this using ordered set. If there is such a $ai$, then increase $xi$ by 1. After the loop the array $xi$ is fully computed, so we can just find the element which brings the maximum number of records, and minimum among such.

一个元素比它前面的所有元素都大，称此元素为$record$。

首先用一个标记数组$r[i]$表示第i个元素是（1）否（0）是$record$，在$O(N)$时间内可以求出。

然后考虑移除一个元素$i$后，能影响哪些$record$？

自身。还有：在他之后的，某些原本不是$record$的，因为$i$的移除，成为$record$的。

这些$record$（称为$j$）有这样的特点：比从1到$j-1$除$i$外的元素都大，但不比$i$大。

记从1到$j-1$除$i$外的元素的最大值为$k$，则可表示为$k<j<=i$。此不等式为本题“题眼”。

这样就有了官方的$O(N^2)$解法：枚举要移除的元素$i$->检查$i$及在他之后的$j$，计算$record$增量->得移除后能使$record$最大的元素。

然而BB半天，$1e5$数据，显然超时。

这时候想想“题眼”，那个神奇的不等式。

同样设（删除$j$后的）增量数组$x[j]$，$O(N)$顺推并维护$k$和$i$不就可以了？

如果$j>i$，说明$j$是$record$，$x[j]--$，并更新$j$为$i$，$k$为$i$；

如果$k<j=<i$，“题眼”，$x[i]++$（注意是$i$），并更新$k$为$j$。

最后的问题，如果存在相同的元素，映射不唯一，一首凉凉？

回头审题，$distinct$，计划通，代码非常简单。

小坑点，第一个元素是不算$record$的。

AC代码

#include<bits/stdc++.h>
using namespace std;

int a[100050];

int main()
{
    int n;
    while(cin>>n){
        memset(a,0,sizeof(a));
        int x,m2nd=0,m1st=0;
        for(int i=1;i<=n;i++){
            cin>>x;
            if(x>m1st){
                m2nd=m1st;
                m1st=x;
                a[x]--;
            }
            else if(x>m2nd){
                m2nd=x;
                a[m1st]++;
            }
        }
        int ma=-0x3f3f3f3f,mi;
        for(int i=1;i<=n;i++)
            if(a[i]>ma)
                ma=a[i],mi=i;
        printf("%d\n",mi);
    }
    return 0;
}