Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final) D. Array Restoration(示例代码

Posted 2020-12-30 lis-

D. Array Restoration

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Initially there was an array $\mathrm{aa consisting\; of nn integers.\; Positions\; in\; it\; are\; numbered\; from 11 to nn.}$

Exactly $\mathrm{qq queries\; were\; performed\; on\; the\; array.\; During\; the ii-th\; query\; some\; segment (li,ri)(li,ri) (1\le li\le ri\le n)(1\le li\le ri\le n) was\; selected\; and\; values\; of\; elements\; on\; positions\; from lili to riri inclusive\; got\; changed\; to ii.\; The\; order\; of\; the\; queries\; couldn‘t\; be\; changed\; and\; all qq queries\; were\; applied.\; It\; is\; also\; known\; that\; every\; position\; from 11 to nn got\; covered\; by\; at\; least\; one\; segment.}$

We could have offered you the problem about checking if some given array (consisting of $\mathrm{nn integers\; with\; values\; from 11 to qq)\; can\; be\; obtained\; by\; the\; aforementioned\; queries.\; However,\; we\; decided\; that\; it\; will\; come\; too\; easy\; for\; you.}$

So the enhancement we introduced to it is the following. Some set of positions (possibly empty) in this array is selected and values of elements on these positions are set to $00.$

Your task is to check if this array can be obtained by the aforementioned queries. Also if it can be obtained then restore this array.

If there are multiple possible arrays then print any of them.

Input

The first line contains two integers $\mathrm{nn and qq (1\le n,q\le 2?1051\le n,q\le 2?105)\; —\; the\; number\; of\; elements\; of\; the\; array\; and\; the\; number\; of\; queries\; perfomed\; on\; it.}$

The second line contains $\mathrm{nn integer\; numbers a1,a2,\dots ,ana1,a2,\dots ,an (0\le ai\le q0\le ai\le q)\; —\; the\; resulting\; array.\; If\; element\; at\; some\; position jj is\; equal\; to 00then\; the\; value\; of\; element\; at\; this\; position\; can\; be\; any\; integer\; from 11 to qq.}$

Output

Print "YES" if the array $\mathrm{aa can\; be\; obtained\; by\; performing qq queries.\; Segments (li,ri)(li,ri) (1\le li\le ri\le n)(1\le li\le ri\le n) are\; chosen\; separately\; for\; each\; query.\; Every\; position\; from 11 to nn should\; be\; covered\; by\; at\; least\; one\; segment.}$

Otherwise print "NO".

If some array can be obtained then print $\mathrm{nn integers\; on\; the\; second\; line\; —\; the ii-th\; number\; should\; be\; equal\; to\; the ii-th\; element\; of\; the\; resulting\; array\; and\; should\; have\; value\; from 11 to qq.\; This\; array\; should\; be\; obtainable\; by\; performing\; exactly qq queries.}$

If there are multiple possible arrays then print any of them.

Examples

input

Copy

4 3
1 0 2 3

output

Copy

YES
1 2 2 3

input

Copy

3 10
10 10 10

output

Copy

YES
10 10 10 

input

Copy

5 6
6 5 6 2 2

output

Copy

NO

input

Copy

3 5
0 0 0

output

Copy

YES
5 4 2

Note

In the first example you can also replace $\mathrm{00 with 11 but\; not\; with 33.}$

In the second example it doesn‘t really matter what segments to choose until query $\mathrm{1010 when\; the\; segment\; is (1,3)(1,3).}$

The third example showcases the fact that the order of queries can‘t be changed, you can‘t firstly set $(1,3)(1,3) to 66 and\; after\; that\; change (2,2)(2,2) to 55.\; The\; segment\; of 55 should\; be\; applied\; before\; segment\; of 66.$

There is a lot of correct resulting arrays for the fourth example.

 

题意：起初我们有一个n个数的排列，然后我们随机让一些数变成了0，这些数的范围都是1-q，然后让我们求原来这个排列，他的排列求来的方法是染色，它总共有i次染色，每次

染色一个区间，让这个区间的数都变成i

 

思路：我们第i次染色会让这个区间的数都变成i，然后我们仔细想想，相同的数是连续的，如果不是连续的，肯定是后面再次在他们中间染了色，所以有规律

两个相同 的数中间肯定都是比它大的数不然就是错的，然后我们可以判断这一段的区间最小值是不是等于它就好，用线段树维护区间最小值，然后我们再来解决0的问题

因为他是随机让一些数变成的0，所以我们可以让0变成1-q中间的任意一个数，很容易发现，我们把0变成两边的一个数就行，就不要在意相同数区间有比他小的数

我们还要注意，因为我们是按顺序染色，所以最后我们肯定有一个数q，最后一次染色留下来的，如果没有这个数，我们要把其中一个0变成q，如果也没有那就直接判断错

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct sss
{
    int left;
    int right;
    int mn;
}a[200001*4];
int n,m;
int c[200001];
int f[200001];
void build(int cnt,int left,int right)
{
    a[cnt].left=left;
    a[cnt].right=right;
    if(left==right)
    {
        a[cnt].mn=c[left];
        return;
    }
    int mid=(left+right)/2;
    build(cnt*2,left,mid);
    build(cnt*2+1,mid+1,right);
    a[cnt].mn=min(a[cnt*2].mn,a[cnt*2+1].mn);
}
int updata(int cnt,int left,int right)
{
    if(left<=a[cnt].left&&right>=a[cnt].right)
    {
        return a[cnt].mn;
    }
    int mid=(a[cnt].left+a[cnt].right)/2;
    if(mid>=right) return updata(cnt*2,left,right);
    else if(left>mid) return updata(cnt*2+1,left,right);
    else{
        return min(updata(cnt*2,left,mid),updata(cnt*2+1,mid+1,right));
    } 
}
int main()
{
    scanf("%d%d",&n,&m);
    int flag1=0,flag2=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&c[i]);
        if(c[i]==m) flag1=1;
        if(c[i]==0) flag2=1;
    }
    if(flag1==0&&flag2==0)
    {
        printf("NO");
        return 0;
    }
    if(flag1==0)
    {
        for(int i=1;i<=n;i++)
        {
            if(c[i]==0)
            {
                c[i]=m;
                break;
            }
        }
    }
    int ans=1;
    for(int i=1;i<=n;i++)
    {
        if(c[i]==0) c[i]=ans;
        else ans=c[i];
    }
    build(1,1,n);
    for(int i=1;i<=n;i++)
    {
        if(f[c[i]]==0) f[c[i]]=i;
        else{
            if(updata(1,f[c[i]],i)==c[i])
            {
                f[c[i]]=i;
                 continue;
            }
            printf("NO");
            return 0;
        }
    }
    printf("YES
");
    for(int i=1;i<=n;i++)
    {
        if(i==1) printf("%d",c[i]);
        else printf(" %d",c[i]);
    }
}

 

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