CF 286C(Main Sequence-贪心括号匹配)

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内容目录
C. Main Sequence
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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.
  • If {a1, a2, ..., al} and {b1, b2, ..., bk} are
    correct bracket sequences, then sequence {a1, a2, ..., al, b1, b2, ..., bk} (their
    concatenation) also is a correct bracket sequence.
  • If {a1, a2, ..., al} —
    is a correct bracket sequence, then sequence  also is a correct bracket sequence, where v (v > 0) is
    an integer.

For example, sequences {1, 1,  - 1, 2,  - 2,  - 1} and {3,  - 3} are
correct bracket sequences, and {2,  - 3} is not.

Moreover, after Vova became a shaman, he learned the most important correct bracket sequence {x1, x2, ..., xn},
consisting of nintegers. As sequence x is the most
important, Vova decided to encrypt it just in case.

Encrypting consists of two sequences. The first sequence {p1, p2, ..., pn} contains
types of brackets, that is, pi = |xi| (1 ≤ i ≤ n).
The second sequence {q1, q2, ..., qt} contains t integers
— 这些地方必须取负数 {x1, x2, ..., xn}.

Unfortunately, Vova forgot the main sequence. But he was lucky enough to keep the encryption: sequences {p1, p2, ..., pn} and{q1, q2, ..., qt}.
Help Vova restore sequence x by the encryption. If there are multiple sequences that correspond to the encryption, restore any of them. If there are no
such sequences, you should tell so.

Input

The first line of the input contains integer n (1 ≤ n ≤ 106).
The second line contains n integers: p1, p2, ..., pn (1 ≤ pi ≤ 109).

The third line contains integer t (0 ≤ t ≤ n),
followed by t distinct integers q1, q2, ..., qt (1 ≤ qi ≤ n).

The numbers in each line are separated by spaces.

Output

Print a single string "NO" (without the quotes) if Vova is mistaken and a suitable sequence {x1, x2, ..., xn} doesn't
exist.

Otherwise, in the first line print "YES" (without the quotes) and in the second line print n integers x1, x2, ..., xn (|xi| = pi; xqj < 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("NOn");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("NOn");
	else
	{
		printf("YESn");
		for (int i=1;i<n;i++) printf("%d ",ans[i]);printf("%dn",ans[n]);
	}

	return 0;
}