Codeforces Beta Round #37 C. Old Berland Language 暴力 dfs

C. Old Berland Language

题目连接：

http://www.codeforces.com/contest/37/problem/C

Description

Berland scientists know that the Old Berland language had exactly n words. Those words had lengths of l1, l2, ..., ln letters. Every word consisted of two letters, 0 and 1. Ancient Berland people spoke quickly and didn’t make pauses between the words, but at the same time they could always understand each other perfectly. It was possible because no word was a prefix of another one. The prefix of a string is considered to be one of its substrings that starts from the initial symbol.

Help the scientists determine whether all the words of the Old Berland language can be reconstructed and if they can, output the words themselves.

Input

The first line contains one integer N (1 ≤ N ≤ 1000) — the number of words in Old Berland language. The second line contains N space-separated integers — the lengths of these words. All the lengths are natural numbers not exceeding 1000.

Output

If there’s no such set of words, in the single line output NO. Otherwise, in the first line output YES, and in the next N lines output the words themselves in the order their lengths were given in the input file. If the answer is not unique, output any.

Sample Input

3
1 2 3

Sample Output

YES
0
10
110

Hint

题意

有n个串，并且每个串的长度为p[i]

你需要构造出这n个串，并且这n个串互相不为其他串的前缀

问你是否能够构造出来

不能输出no

题解：

直接dfs一波，就莽过去了……

复杂度很显然是最多往下面搜1000层吧……

感觉分支不是很多的样子

233

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n;
struct node
{
    int x,y;
    string ans;
}p[maxn];
int now,ok;
bool cmp(node a,node b)
{
    return a.x<b.x;
}
bool cmp2(node a,node b)
{
    return a.y<b.y;
}
void dfs(int x,string s)
{
    if(x==p[now].x)
    {
        p[now++].ans=s;
        if(now==n)ok=1;
        return;
    }
    dfs(x+1,s+'0');
    if(now==n)return;
    dfs(x+1,s+'1');
    if(now==n)return;
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&p[i].x);
        p[i].y=i;
        p[i].ans="";
    }
    sort(p,p+n,cmp);
    dfs(0,"");
    if(!ok)printf("NO
");
    else{
        printf("YES
");
        sort(p,p+n,cmp2);
        for(int i=0;i<n;i++)
            cout<<p[i].ans<<endl;
    }
}