[ACM] zoj 3818 Pretty Poem (2014 ACMICPC Regional 牡丹江站网络赛 J题）

Pretty Poem

More technically, we call a poem pretty if it can be decomposed into one of the following rhyme scheme: "ABABA" or "ABABCAB". The symbol A, B and C are
different continuous non-empty substrings of the poem. By the way, punctuation characters should be ignored when considering the rhyme scheme.

You are given a line of poem, please determine whether it is pretty or not.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a line of poem S (1 <= length(S) <= 50). S will only contains alphabet characters or punctuation characters.

Output

For each test case, output "Yes" if the poem is pretty, or "No" if not.

Sample Input

3
niconiconi~
pettan,pettan,tsurupettan
wafuwafu

Sample Output

Yes
Yes
No

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

解题思路：

题意为输入一个只包含字母和标点符号的字符串，问该字符串是否符合 "ABABA" or "ABABCAB"的形式，其中A,B,C为原字符串中连续的不相同的子串。

比如niconiconi 符合ABABA的形式，因为 A= ni  B= con， 判断的时候要忽略掉字符串中的标点符号

思路为首先提取出来字符串中的字母，然后分别判断是否符合以上两种形式，判断ABABA的时候枚举A的长度，那么B的长度也就确定了，判断ABABCAB的时候，枚举A,B的长度，那么C的长度也就确定了。做题中出现的问题是char
s[60],输入字符串的时候用了cin>>s,一直WA，换了gets(s)以后就过了，难道测试数据中字符串包含空格？可是空格不是标点符号啊。。。

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stack>
#include <queue>
#include <iomanip>
#include <cmath>
#include <string.h>
using namespace std;
#define ll long long
const int inf=0x3f3f3f3f;
int n;

int main()
{
    cin>>n;
    getchar();
    while(n--)
    {
        char s[60];
        char temp[60];
        gets(s);
        int l=strlen(s);
        if(l<5)
        {
            cout<<"No"<<endl;
            continue;
        }
        int len=0;
        for(int i=0;i<l;i++)
        {
            if((s[i]>=65&&s[i]<=90)||(s[i]>=97&&s[i]<=122))
            {
                temp[len]=s[i];
                len++;
            }
        }
        temp[len]='\0';
        if(len<5)
        {
            cout<<"No"<<endl;
            continue;
        }
        //ABABA的情况
        int la=0,lb=0;//A的长度，B的长度
        bool ok;
        for(la=1;;la++)
        {
            ok=0;
            if(len-3*la<2)
                break;
            if((len-3*la)%2!=0)
                continue;
            lb=(len-3*la)/2;
            string A="";
            string B="";
            for(int i=0;i<la;i++)
                A+=temp[i];
            for(int i=la;i<la+lb;i++)
                B+=temp[i];
            if(A==B)
                continue;

            string ans="";
            ans=A+B+A+B+A;
            int i;
            for(i=0;i<len&&ans[i]==temp[i];i++){}
            if(i==len)
            {
                ok=1;
                break;
            }
        }
        if(ok)
        {
            cout<<"Yes"<<endl;
            continue;
        }
        //ABABCAB的情况
        int lc=0;
        for(la=1;la<len-3;la++)
        {
            ok=0;
            for(lb=1;lb<len-3;lb++)
            {
                if(3*la+3*lb>=len)
                    break;
                if(len-3*la-3*lb<0)
                    break;
                lc=len-3*la-3*lb;
                string A="";
                string B="";
                string C="";
                for(int i=0;i<la;i++)
                    A+=temp[i];
                for(int i=la;i<la+lb;i++)
                    B+=temp[i];
                for(int i=(la+lb)*2;i<(la+lb)*2+lc;i++)
                    C+=temp[i];
                if(A==B||A==C||B==C)//注意这一点，ABC不能相同
                    continue;
                string ans="";
                ans=A+B+A+B+C+A+B;
                int i;
                for(i=0;i<len&&ans[i]==temp[i];i++){}
                if(i==len)
                {
                    ok=1;
                    break;
                }
            }
            if(ok)
                break;
        }
        if(ok)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}