Sort the Array

最新推荐文章于 2022-08-16 18:27:16 发布

原创最新推荐文章于 2022-08-16 18:27:16 发布 · 372 阅读

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本文探讨了一个关于通过反转数组中某一段来实现整个数组升序排列的问题。文章首先介绍了问题背景及输入输出要求，随后详细解释了算法思路，并提供了一段C++代码实现。通过对数组的遍历和特定条件下的反转操作，判断是否能通过一次反转达到排序目的。

原题链接
B. Sort the Array
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of n distinct integers.

Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the array a (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.

Input
The first line of the input contains an integer n (1 ≤ n ≤ 105) — the size of array a.

The second line contains n distinct space-separated integers: a[1], a[2], …, a[n] (1 ≤ a[i] ≤ 109).

Output
Print “yes” or “no” (without quotes), depending on the answer.

If your answer is “yes”, then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.

Examples
input
3
3 2 1
output
yes
1 3
input
4
2 1 3 4
output
yes
1 2
input
4
3 1 2 4
output
no
input
2
1 2
output
yes
1 1
Note
Sample 1. You can reverse the entire array to get [1, 2, 3], which is sorted.

Sample 3. No segment can be reversed such that the array will be sorted.

Definitions

A segment [l, r] of array a is the sequence a[l], a[l + 1], …, a[r].

If you have an array a of size n and you reverse its segment [l, r], the array will become:

a[1], a[2], …, a[l - 2], a[l - 1], a[r], a[r - 1], …, a[l + 1], a[l], a[r + 1], a[r + 2], …, a[n - 1], a[n].

这个题就是把第一个不是上升的区间找出来，将它反转后若是发现还不满足那么他绝对就不可以反转，若是循环结束后没有出现问题那么就说明它是可以的

#include <cstdio>
#include <algorithm>
using namespace std;
int a[100010],f[100010],p=0,q=0;//f[i]是未来的复制数组（a[0]=0,0代表升，1代表降），p是第一个f为1的位置的下标，q是第一个1转0的位置的下标
int main(){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
                scanf("%d",&a[i]);
        a[n]=1000000007;//因为要借助它找到最后一个q
        for(int i=1;i<=n;i++){
                if(a[i]-a[i-1]<0){
                        p=i;
                        for(int j=p+1;j<=n;j++){
                                if(a[j]-a[j-1]>0){
                                        q=j;
                                        goto END;
                                }
                        }
                }
        }
        END:;
        if(p==0 && q==0){//说明此时是完全递增的，不需要修改，那么任取一个值
                printf("yes\n%d %d\n",1,1);
        }
        else{//那么这时pq必然有值，只需要反转a[p-1]到a[q-1]这一段即可
                for(int i=0;i<(p-1);i++) f[i]=a[i];
                for(int i=q-1,t=0;i>=(p-1);i--,t++) f[p-1+t]=a[i];
                for(int i=q;i<n;i++) f[i]=a[i];
                bool flag=true;
                for(int i=1;i<n;i++){
                        if(f[i]-f[i-1]<0){
                                flag=false;
                                break;
                        }
                }
                if(flag){
                        printf("yes\n");
                        printf("%d %d\n",p,q);
                        //for(int i=0;i<n;i++) printf("%d ",f[i]);
                }
                else printf("no\n");
        }
        return 0;
}