2435: Dictionary Order

 2435: Dictionary Order

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	32768K	825	146	Standard

One day, skywind will build a new dictionary includes a lot of english words. In this dictionary, one word is standing in the front of other word if these conform to the following principles: 1. The shorter word is better and ahead. 2. From letter 'a' to 'z', Let xi and yi is the first place of letter 'a' or 'A' in each word (it is infinity if this letter can't be find), the smaller is better. If they are same, continue the next letter until 'z' or 'Z'. 3. Compare using common dictionary order if they are not comparable as above.

Input

The first line of each case is a integer n (n is less than 1000) who is the number of words. The input is finished if n is zero. There is just one word in the next n lines. The word consists of only lowercase and uppercase letter, its length is no more than 20.

Output

Output the dictionary order by the above rules. Print one word in one line. Each case should be separated from the previous by one blank line.

Sample Input

5
Abovee
Sample
Input
AND
Output
0

Sample Output

AND
Input
Abovee
Sample
Output

 

Problem Source: provided by skywind

 

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This problem is used for contest: 94 

 

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    char s[21];
}str;

int pos(char s[], char a) {
    int pos = 50;
    for(int i = 0; s[i] != '/0'; i++){
        if(s[i] == a){
            pos = i;
            break;
        }
    }
    return pos;
}

int cmp ( const void *a , const void *b ) {
    if( strlen((*(str *)a).s) != strlen((*(str *)b).s) )
        return strlen((*(str *)a).s) - strlen((*(str *)b).s);
    else {
        char s1[21], s2[21];
        int i;
        strcpy(s1, (*(str *)a).s);
        strcpy(s2, (*(str *)b).s);
        for(i = 0; i < strlen(s1); i++){
            if(s1[i] >= 'A' && s1[i] <= 'Z')
                s1[i] = s1[i] - 'A' + 'a';
        }
        for(i = 0; i < strlen(s2); i++){
            if(s2[i] >= 'A' && s2[i] <= 'Z')
                s2[i] = s2[i] - 'A' + 'a';
        }
        for(i = 'a'; i <= 'z'; i++){
            if(pos(s1, i) < pos(s2, i)){
                return -1;
              //  break;
            }
            if(pos(s1, i) > pos(s2, i)){
                return 1;
             //   break;
            }
        }
        if(i == 'z' + 1){
            if( strcmp((*(str *)a).s, (*(str *)b).s) >0 )
                return 1;
            else
                return 0;
        }
    }
}

int main () {
    int n,i;
    str ss[1000];

    scanf("%d", &n);
    if(n == 0)
        goto end;
    for(i = 0; i < n; i++)
        scanf("%s", ss[i].s);
    qsort(ss, n, sizeof(ss[0]), cmp);
    for(i = 0; i < n; i++)
        printf("%s/n", ss[i].s);

    while(scanf("%d",&n) && n != 0){
        for(i = 0; i < n; i++)
            scanf("%s", ss[i].s);
        qsort(ss, n, sizeof(ss[0]), cmp);
          printf("/n");
        for(i = 0; i < n; i++)
            printf("%s/n", ss[i].s);
    }
    end:;
    return 0;
}

 

 

 

#include<iostream>
#include<string>
#include<algorithm>
#include<cctype>
using namespace std;
struct node
{
    string str,s;
    int size;
};
node a[1001];
int pos(string s,char ch)
{
    int p=50;
    for(int i=0;i<s.size();i++)  if(s[i]==ch) {  p=i;  break;}
    return p;
}
bool cmp(node x,node y)
{
   if(x.size==y.size)
   {
       string s1=x.s,s2=y.s;int i;
       for( i='a';i<='z';i++)
       {
           if(pos(s1,i)!=pos(s2,i))
           {
                return pos(s1,i)<pos(s2,i);
           }
       }
       if(i=='z'+1) return x.str<y.str;
   }
   return x.size<y.size;
}
int main()
{
    int n,i,j;
    int f=0;
    while(scanf("%d",&n)==1&&n)
    {
        if(f==0) f=1;else printf("/n");
        for(int i=0;i<n;i++)
        {
            cin>>a[i].str;a[i].s=a[i].str;
            a[i].size=a[i].str.size();
            for(int j=0;j<a[i].size;j++) a[i].s[j]=tolower(a[i].s[j]);
        }
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++) cout<<a[i].str<<endl;
    }
    return 0;
}