CF1455D Sequence and Swaps

Description

You are given a sequence $ a $ consisting of $ n $ integers $ a_1, a_2, dots, a_n $ , and an integer $ x $ . Your task is to make the sequence $ a $ sorted (it is considered sorted if the condition $ a_1 le a_2 le a_3 le dots le a_n $ holds).

To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer $ i $ such that $ 1 le i le n $ and $ a_i > x $ , and swap the values of $ a_i $ and $ x $ .

For example, if $ a = [0, 2, 3, 5, 4] $ , $ x = 1 $ , the following sequence of operations is possible:

1. choose $ i = 2 $ (it is possible since $ a_2 > x $ ), then $ a = [0, 1, 3, 5, 4] $ , $ x = 2 $ ;
2. choose $ i = 3 $ (it is possible since $ a_3 > x $ ), then $ a = [0, 1, 2, 5, 4] $ , $ x = 3 $ ;
3. choose $ i = 4 $ (it is possible since $ a_4 > x $ ), then $ a = [0, 1, 2, 3, 4] $ , $ x = 5 $ .

Calculate the minimum number of operations you have to perform so that $ a $ becomes sorted, or report that it is impossible.

Solution

假设已经把$x$加入数列，对这个数列排序会得到最终的结果

发现将$x$插入数列相当于将数列中的一端向右移动一段，从前到后扫数组，当新数组小于原数组对应位时选择交换，每扫一位检查序列单调性

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
int n,x,a[505],sav[505],ans;
bool tag;
inline int read()
{
    int f=1,w=0;
    char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') w=(w<<1)+(w<<3)+ch-'0',ch=getchar();
    return f*w;
}
bool check(int pos)
{
    bool ret=true;
    for(int i=pos;i<n;i++) if(a[i]>a[i+1]) ret=false;
    return ret;
}
int main()
{
    for(int t=read();t;t--)
    {
        n=read(),x=read(),tag=false,ans=0;
        for(int i=1;i<=n;i++) sav[i]=a[i]=read();
        for(int i=1;i<n;i++) if(a[i]>a[i+1]) tag=true;
        if(!tag)
        {
            puts("0");continue;
        }
        sav[n+1]=x,sort(sav+1,sav+n+2),tag=false;
        for(int i=1;i<=n;i++)
        {
            if(a[i]!=sav[i])
                if(a[i]>sav[i]) swap(x,a[i]),++ans;
                else tag=true;
            if(a[i]<a[i-1]) tag=true;
            if(check(i)) break;
        }
        if(tag) puts("-1");
        else printf("%d
",ans);
    }
    return 0;
}