zoukankan      html  css  js  c++  java
  • 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 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 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 nn (1n1001≤n≤100) — the number of strings.

    The next 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 nn given strings in required order, print "NO" (without quotes).

    Otherwise print "YES" (without quotes) and 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:

     1  for (int i=0; i<=word[p].len-word[p-1].len; i++)//每个头位置的指针
     2     {
     3         flag=0;
     4         pi=i;//检查在I为头的情况下的是否匹配的指针
     5         for(int j=0; j<word[p-1].len;j++)
     6         {
     7             if(word[p].sub[pi]!=word[p-1].sub[j])
     8             {
     9                 flag=1;//如果有不同
    10                 break;
    11             }
    12             else
    13             {
    14                 pi++;//继续匹配
    15                 continue;
    16             }
    17         }
    18         if (flag==0)break;
    19     }

    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;
    }
    有不懂欢迎咨询 QQ:1326487164(添加时记得备注)
  • 相关阅读:
    JS实现继承的几种方式
    跨平台APP----对Cordova,APPCan,DCloud,APICloud四大平台的分析
    cordova生成的android项目导入到Android studio 2.X 中遇到的问题解决方案
    链操作相关命令(包括启动,重启,删除)
    冷钱包和热钱包有什么区别?
    常用命令之git/linux
    centos安装git,go,shasum,okexchain环境
    iterm2的下载安装与配置
    使用jsdoc-to-markdown提前js文件的文档
    基于sphinx的文档(一)将md转为rst
  • 原文地址:https://www.cnblogs.com/bluefly-hrbust/p/9130359.html
Copyright © 2011-2022 走看看