[ABC274D] Robot Arms 2

Problem Statement

You are given a sequence $A = (A_1, A_2, \dots, A_N)$ of length $N$ consisting of positive integers, and integers $x$ and $y$.
Determine whether it is possible to place $N+1$ points $p_1, p_2, \dots, p_N, p_{N+1}$ in the $xy$-coordinate plane to satisfy all of the following conditions. (It is allowed to place two or more points at the same coordinates.)

• $p_1 = (0, 0)$.
• $p_2 = (A_1, 0)$.
• $p_{N+1} = (x, y)$.
• The distance between the points $p_i$ and $p_{i+1}$ is $A_i$. ($1 \leq i \leq N$)
• The segments $p_i p_{i+1}$ and $p_{i+1} p_{i+2}$ form a $90$ degree angle. ($1 \leq i \leq N - 1$)

Constraints

• $2 \leq N \leq 10^3$
• $1 \leq A_i \leq 10$
• $|x|, |y| \leq 10^4$
• All values in the input are integers.

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Input

The input is given from Standard Input in the following format:

$N$ $x$ $y$
$A_1$ $A_2$ $\dots$ $A_N$

Output

If it is possible to place $p_1, p_2, \dots, p_N, p_{N+1}$ to satisfy all of the conditions in the Problem Statement, print Yes; otherwise, print No.

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Sample Input 1

3 -1 1
2 1 3

Sample Output 1

Yes

The figure below shows a placement where $p_1 = (0, 0)$, $p_2 = (2, 0)$, $p_3 = (2, 1)$, and $p_4 = (-1, 1)$. All conditions in the Problem Statement are satisfied.

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Sample Input 2

5 2 0
2 2 2 2 2

Sample Output 2

Yes

Letting $p_1 = (0, 0)$, $p_2 = (2, 0)$, $p_3 = (2, 2)$, $p_4 = (0, 2)$, $p_5 = (0, 0)$, and $p_6 = (2, 0)$ satisfies all the conditions. Note that multiple points may be placed at the same coordinates.

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Sample Input 3

4 5 5
1 2 3 4

Sample Output 3

No

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Sample Input 4

3 2 7
2 7 4

Sample Output 4

No

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Sample Input 5

10 8 -7
6 10 4 1 5 9 8 6 5 1

Sample Output 5

Yes

首先考虑将 \(x\) 坐标和 \(y\) 坐标分开考虑，很明显他们是互不干扰的。然后可以用一个类似背包的方法计算。负数存在数组时可以通过加一个值计算。其实此题应该是可以 bitset 优化的（不过我没写）。注意 \(p2=(A1,0)\)

#include<bits/stdc++.h> 
const int N=1005,M=2e4+5,T=1e4;
int n,x,y,dp[N][M],a[N],b[N],p,q,u;
int solve(int a[],int n,int x,int t)
{
	memset(dp,0,sizeof(dp));
	int st;
	if(!t)
		dp[1][a[1]+T]=1,st=2;
	else
		dp[0][T]=1,st=1;
	for(int i=st;i<=n;i++)
	{
		for(int j=0;j<M;j++)
		{
			if(j>=a[i])
				dp[i][j]=dp[i-1][j-a[i]];
			if(j+a[i]<M)
				dp[i][j]|=dp[i-1][j+a[i]];
		}
	}
//	printf("%d\n",dp[n][x]);
	return dp[n][T+x];
}
int main()
{
	scanf("%d%d%d",&n,&x,&y);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&u);
		if(i&1)
			a[++p]=u;
		else
			b[++q]=u;
	}
	if(solve(a,p,x,0)&&solve(b,q,y,1))
		printf("Yes");
	else
		printf("No");
}