CF 286C(Main Sequence-贪心括号匹配)

Posted 2022-10-25

C. Main Sequence

time limit per test

memory limit per test

input

output

As you know, Vova has recently become a new shaman in the city of Ultima Thule. So, he has received the shaman knowledge about the correct bracket sequences. The shamans of Ultima Thule have been using lots of different types of brackets since prehistoric times. A bracket type is a positive integer. The shamans define a correct bracket sequence as follows:

• An empty sequence is a correct bracket sequence.
• a₁,a₂, ...,a_l and b₁,b₂, ...,b_k are correct bracket sequences, then sequence a₁,a₂, ...,a_l,b₁,b₂, ...,b_k
• a₁,a₂, ...,a_l — is a correct bracket sequence, then sequence



• also is a correct bracket sequence, wherev(v> 0)

1, 1,  - 1, 2,  - 2,  - 1 and 3,  - 3 are correct bracket sequences, and 2,  - 3

most important correct bracket sequence x₁, x₂, ..., x_n, consisting of nintegers. As sequence x

p₁, p₂, ..., p_n contains types of brackets, that is, p_i = |x_i| (1 ≤ i ≤ n). The second sequence q₁, q₂, ..., q_t contains t integers — 这些地方必须取负数 x₁, x₂, ..., x_n.

p₁, p₂, ..., p_n andq₁, q₂, ..., q_t. Help Vova restore sequence x

Input

n (1 ≤ n ≤ 10⁶). The second line contains n integers: p₁, p₂, ..., p_n (1 ≤ p_i ≤ 10⁹).

t (0 ≤ t ≤ n), followed by t distinct integers q₁, q₂, ..., q_t (1 ≤ q_i ≤ n).

The numbers in each line are separated by spaces.

Output

NO" (without the quotes) if Vova is mistaken and a suitable sequence x₁, x₂, ..., x_n

YES" (without the quotes) and in the second line print n integers x₁, x₂, ..., x_n (|x_i| = p_i; x_qj < 0). If there are multiple sequences that correspond to the encrypting, you are allowed to print any of them.

Sample test(s)

input

2 1 1 0

output

YES 1 -1

input

4 1 1 1 1 1 3

output

YES 1 1 -1 -1

input

3 1 1 1 0

output

NO

input

4 1 2 2 1 2 3 4

output

YES 1 2 -2 -1

贪心，从后向前做，尽可能取左括号。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<functional>
#include<algorithm>
#include<stack>
using namespace std;
#define MAXN (1000000+10)
#define MAXP (1000000000+10)
int n,m,a[MAXN],ans[MAXN],tot;
bool b[MAXN];
int s[MAXN],size=0;
int main()

  memset(b,0,sizeof(b));
  scanf("%d",&n);tot=n+1;
  if (n%2) printf("NO\\n");return 0;
  for (int i=1;i<=n;i++) scanf("%d",&a[i]);
  scanf("%d",&m);
  for (int i=1;i<=m;i++) 
  
    int p;
    scanf("%d",&p);
    b[p]=1;
  
  for (int i=n;i;i--)
  
    if (size==0||s[size]!=a[i]||b[i])
    
      s[++size]=a[i];ans[--tot]=-a[i];
    
    else ans[--tot]=s[size--];
  
  if (size) printf("NO\\n");
  else
  
    printf("YES\\n");
    for (int i=1;i<n;i++) printf("%d ",ans[i]);printf("%d\\n",ans[n]);
  
  
  return 0;