CF 429C

 

Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.

Iahub asks Iahubina: can you build a rooted tree, such that

• each internal node (a node with at least one son) has at least two sons;
• node i has ci nodes in its subtree?

Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.

Input

The first line of the input contains integer n (1 ≤ n ≤ 24). Next line contains n positive integers: the i-th number represents ci(1 ≤ ci ≤ n).

Output

Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).

Sample test(s)

input

4
1 1 1 4

output

YES

input

5
1 1 5 2 1

output

NO

 

 

中文题意：

给出一棵树中每个节点子树的大小，要求每个非叶节点至少有2个儿子，问是否能构造出这样一棵树

 

解法：

N<=24，因此可以用状压解。

首先可以证明如果叶节点<N/2，那么肯定无解

所以我们可以把非叶节点和叶节点分开处理。

我们每次其实就是要往一个非叶点里塞点，或者说给每个点找父节点

我们设状态 f[i][j][k]为已塞完前i个非叶节点，非叶节点是否被塞过（即是否已有父节点的状态为j，还有k个叶节点没有父节点

然后对于每个点，枚举塞哪些非叶节点即可

 

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

bool f[14][8201][14];
int zt[9011],num[9011];
int a[101];
int n,i,j,k,cnt,t;

void Work(int x,int y,int z)
{
    int lef,i,yf;
    for(i=0;i<(1<<t);i++)if(!(y&i)){
        lef=a[x]-1-zt[i];
        if(lef>=0&&lef<=z){
            if(lef+num[i]>1)f[x][y|i][z-lef]=true;
        }
    }
}

int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&k);
        if(k==1)cnt++;
        else a[++t]=k;
    }
    for(i=1;i<t;i++)
        for(j=i+1;j<=t;j++)if(a[i]<a[j])swap(a[i],a[j]);
    if(n==1&&cnt==1)printf("YES\n");
    else if(cnt<n/2||cnt==n)printf("NO\n");
    else{
        for(i=0;i<(1<<t);i++){
            for(j=1;j<=t;j++)if((i&(1<<(j-1)))){
                zt[i]+=a[j];
                num[i]++;
            }
        }
        f[0][1][cnt]=true;
        for(i=0;i<t;i++)
            for(j=0;j<(1<<t);j++)
                for(k=0;k<=cnt;k++)if(f[i][j][k])Work(i+1,j,k);
        if(f[t][(1<<t)-1][0])printf("YES\n");
        else printf("NO\n");
    }
}