Codeforces Round #486 (Div. 3)-B. Substrings Sort

B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{nn strings.\; Each\; string\; consists\; of\; lowercase\; English\; letters.\; Rearrange\; (reorder)\; the\; given\; strings\; in\; such\; a\; way\; that\; for\; every\; string,\; all\; strings\; that\; are\; placed\; before\; it\; are\; its substrings.}$

String $\mathrm{aa is\; a substring of\; string bb if\; it\; is\; possible\; to\; choose\; several consecutive letters\; in bb in\; such\; a\; way\; that\; they\; form aa.\; For\; example,\; string\; "for"\; is\; contained\; as\; a substring in\; strings\; "codeforces",\; "for"\; and\; "therefore",\; but\; is\; not\; contained\; as\; a substring in\; strings\; "four",\; "fofo"\; and\; "rof".}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 1001\le n\le 100)\; —\; the\; number\; of\; strings.}$

The next $\mathrm{nn lines\; contain\; the\; given\; strings.\; The\; number\; of\; letters\; in\; each\; string\; is\; from 11 to 100100,\; inclusive.\; Each\; string\; consists\; of\; lowercase\; English\; letters.}$

Some strings might be equal.

Output

If it is impossible to reorder $\mathrm{nn given\; strings\; in\; required\; order,\; print\; "NO"\; (without\; quotes).}$

Otherwise print "YES" (without quotes) and $\mathrm{nn given\; strings\; in\; required\; order.}$

Examples

input

Copy

5
a
aba
abacaba
ba
aba

output

Copy

YES
a
ba
aba
aba
abacaba

input

Copy

5
a
abacaba
ba
aba
abab

output

Copy

NO

input

Copy

3
qwerty
qwerty
qwerty

output

Copy

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

被HACK掉了。。。只能说数据太水了。。。发现自己的程序问题还这么大，还是太菜了。

本题题意就是是否可以把所给的N个序列摆放成上面的序列是下面序列的子序列

开始想了半天。。。怎么找啊。。。这™序列这么多，要我一个一个比？后来发现，你只需要按照从小到大的序列长度排序就好了啊。。。然后从下往上，两个判断是否上面的那么个序列是下面序列的子序列。o(n)操作嘛，小case.

被HACK掉原因就是我匹配那里写错了

正确写匹配姿势:

 for (int i=0; i<=word[p].len-word[p-1].len; i++)//每个头位置的指针
    {
        flag=0;
        pi=i;//检查在I为头的情况下的是否匹配的指针
        for(int j=0; j<word[p-1].len;j++)
        {
            if(word[p].sub[pi]!=word[p-1].sub[j])
            {
                flag=1;//如果有不同
                break;
            }
            else
            {
                pi++;//继续匹配
                continue;
            }
        }
        if (flag==0)break;
    }

emmmmm最后AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Node
{
    char sub[105];
    int len;
} word[105];
bool cmp(Node x,Node y)
{
    return x.len<y.len;//按照长度排序
}
int pp(int p)
{
    int flag;
    int pi;
    for (int i=0; i<=word[p].len-word[p-1].len; i++)
    {
        flag=0;
        pi=i;
        for(int j=0; j<word[p-1].len;j++)
        {
            if(word[p].sub[pi]!=word[p-1].sub[j])
            {
                flag=1;
                break;
            }
            else
            {
                pi++;
                continue;
            }
        }
        if (flag==0)break;
    }
    if (flag==0)return 0;
    else return 1;
}
int main()
{
    int n;
    int lens;
    int flag;
    while (~scanf("%d",&n))
    {
        flag=0;
        for (int i=0; i<n; i++)
        {
            scanf("%s",word[i].sub);
            lens=strlen(word[i].sub);
            word[i].len=lens;
        }
        sort(word,word+n,cmp);
        for (int i=n-1; i>0; i--)
        {
            if(pp(i)!=0){
            flag=1;
            break;
            }
        }
        if(flag==1)printf("NO
");
        else
        {
            printf("YES
");
            for (int i=0; i<n; i++)
            {
                printf("%s
",word[i].sub);
            }
        }
    }
    return 0;
}