ZOJ 3818

来源：互联网发布：羽毛球教学软件编辑：程序博客网时间：2024/06/01 18:27

这道题没什么固定算法，关键就是思维是否够缜密。

我的方法是先找出DDE模式（其中D的长度必须大于2），再根据具体的情况进行下一步处理。这个具体情况就是当D>E时，E就是A；当D<E时，把E拆成CD，也就是CAB。大的方向说完了，就是细节问题，当D>E时，E能否和D的前半段匹配，如果可以匹配那么D拆成AB后AB是否符合题目要求，当D<E时，D能否和E的后半段匹配，同样E拆成CAB后是否符合题目要求。

最后感谢一下Kewowlo博客里的数据。

代码（C++）：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#define MAX 55using namespace std;char s[MAX];int main(){    //freopen("in.txt","r",stdin);    int t,i,j,k,id,len;    bool flag;    string str,str1,str2,str3;    scanf("%d",&t);    while(t--)    {        scanf("%s",s);        str="";        for(i=0;i<strlen(s);i++)    //去除标点符号        {            if(s[i]>='a'&&s[i]<='z'||s[i]>='A'&&s[i]<='Z') str+=s[i];        }        flag=false;        len=str.length();        for(i=1;i<len;i++)        {            if(str[0]==str[i])            {                if(i<2) continue;   //如果D的长度小于2，直接跳过                if(i*2-1<len&&str.substr(0,i)==str.substr(i,i))   //判断是否存在DD的情况                {                    id=2*i;                    if(len-id>i){   //针对ABABCAB模式                        for(j=i-1,k=0;j>=0;j--,k++)     //判断E的后半段是否有D                        {                            if(str[len-1-k]!=str[j]) break;                        }                        if(j<0){                            str3=str.substr(id,len-i-id);   //判断是否能分成符合条件的ABC，即判断ABC是否不相同且不为空                            for(j=1;j<i;j++)    //枚举A的长度                            {                                str1=str.substr(0,j);                                str2=str.substr(j,i-j);                                if(str1!=str2&&str1!=str3&&str2!=str3) break;                            }                            if(j<i){                               flag=true;                               break;                            }                        }                    }else if(len-id<i&&len>id){  //针对ABABA模式                        for(j=0;j<len-id;j++)   //判断E是否为D的前半段                        {                            if(str[j]!=str[id+j]) break;                        }                        if(j==len-id&&str.substr(0,j)!=str.substr(j,i-j)){  //此时E为A，判断是否D所包含的AB不相同                            flag=true;                            break;                        }                    }                }            }        }        if(flag) printf("Yes\n");        else printf("No\n");    }    return 0;}

题目（http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5350）：

Pretty Poem

Time Limit: 2 Seconds Memory Limit: 65536 KB

Poetry is a form of literature that uses aesthetic and rhythmic qualities of language. There are many famous poets in the contemporary era. It is said that a few ACM-ICPC contestants can even write poetic code. Some poems has a strict rhyme scheme like "ABABA" or "ABABCAB". For example, "niconiconi" is composed of a rhyme scheme "ABABA" with A = "ni" and B = "co".

More technically, we call a poem pretty if it can be decomposed into one of the following rhyme scheme: "ABABA" or "ABABCAB". The symbolA, 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

3niconiconi~pettan,pettan,tsurupettanwafuwafu

Sample Output

YesYesNo

0 0