Securing the Barn poj

                                                                                                                         Securing the Barn

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1860 Accepted: 1191

Description

Farmer John has installed a new security system on the barn and now must issue a valid password to the cows in the herd. A valid password consists of L (3 <= L <= 15) different lower-case characters (from the traditional latin character set 'a'...'z'), has at least one vowel ('a', 'e', 'i', 'o', or 'u'), at least two consonants (non-vowels), and has characters that appear in alphabetical order (i.e., 'abc' is valid; 'bac' is not).

Given a desired length L along with C lower-case characters, write a program to print all the valid passwords of length L that can be formed from those letters. The passwords must be printed in alphabetical order, one per line.

Input

* Line 1: Two space-separated integers, L and C

* Line 2: C space-separated lower-case characters that are the set of characters from which to build the passwords

Output

* Lines 1..?: Each output line contains a word of length L characters (and no spaces). The output lines must appear in alphabetical order.

Sample Input

4 6a t c i s w

Sample Output

acisacitaciwacstacswactwaistaiswaitwastwcistciswcitwistw

Hint

INPUT DETAILS:
Passwords of length 4 chosen from the given six characters

Source

USACO 2005 November Bronze

从六个里选四个字母 需至少1个元音字母，两个辅音字母  然后进行全排列

这道题没什么好解释的~

code:

#include<iostream>#include<algorithm>using namespace std;int num,l,yz,fz,index,len;char a[27];int vis[27];void dfs(int index,int len,int yz,int fz){    if (len==num&&yz>=1&&fz>=2)      {        for (int i=0; i<l; i++)          if (vis[i])             cout<<a[i];                cout<<endl;        return ;    }    for (int i=index; i<l; i++)    {                                            if (a[i]=='a'||a[i]=='e'||a[i]=='i'||a[i]=='o'||a[i]=='u')            {             vis[i]=1;     dfs(i+1,len+1,yz+1,fz);     vis[i]=0;            }        else            {            vis[i]=1;            dfs(i+1,len+1,yz,fz+1);             vis[i]=0;             }                           }}int main(){    cin>>num>>l;    for (int i=0; i<l; i++)    {        cin>>a[i];        vis[i]=0;    }    sort(a,a+l);       dfs(0,0,0,0);       return 0;}