HDOJ 5311 Hidden String(枚举)

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1715    Accepted Submission(s): 604

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n . He wants to find three nonoverlapping substrings s[l1..r1] , s[l2..r2] , s[l3..r3] that:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n

2. The concatenation of s[l1..r1] , s[l2..r2] , s[l3..r3] is "anniversary".

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤100) , indicating the number of test cases. For each test case:

There's a line containing a string s (1≤|s|≤100) consisting of lowercase English letters.

 

Output

For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).

 

Sample Input

   
   
   
   

    
    
    
    2 annivddfdersewwefary nniversarya
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES NO
   
   
   
   

 

题意：在给出的串中是否存在三个子串可以连接成anniversary，且这三个串在给出的串中的位置是依序的不存在重叠。

一直想着KMP怎么搞，原来给出串长度才100，暴力就行了。先在anniversary串中枚举三个子串，再在给出的串中按序查找是否存在这些串就行了，简直水题，我真是智障。。。

代码如下：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
string s="anniversary"; 
string str,s1,s2,s3;
int len_s,len_str; 
int main()
{
	int sign,t,i,j,k;
	scanf("%d",&t);
	while(t--)
	{
		cin>>str;
		len_s=s.length();
		len_str=str.length();
		sign=0;
		for(i=1;i<=len_s-2;++i)//i表示第一个串的长度 
		{
			for(j=1;j<=len_s-1-i;++j)//j表示第二个串的长度 
			{
				k=len_s-i-j;//k表示第三个串的长度 
				s1=s.substr(0,i);//从0位开始往后截取i长度的串 
				s2=s.substr(i,j);
				s3=s.substr(j+i,k);
				int st1=str.find(s1,0);//从第0位开始找s1，找到返回s1串存在的地址，找不到返回-1 
				if(st1==-1)//第一个串错误了，直接跳出确定第二个串的循环 
					break; 
				int st2=str.find(s2,st1+i);
				if(st2==-1)
					continue;
				int st3=str.find(s3,st2+j);
				if(st3==-1)
					continue;
				sign=1;
				printf("YES\n");
				break;
			}
			if(sign)
				break;
		}
		if(!sign)
			printf("NO\n");
	}
	return 0;
}