hdoj--1034--Hidden String（dfs）

Hidden String

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

Total Submission(s): 1677    Accepted Submission(s): 589

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

 

Source

BestCoder 1st Anniversary ($)

 

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#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[1010],op[13]={"anniversary"};
int flog,l,L;
void dfs(int b1,int b2,int num)
{
	int i;
	if(b2>=11&&num<=3)
	{
		flog=1;
		return ;
	}
	if(num>3)
	return ;
	if(b1>=l||flog)
	return ;
	for(i=b1;i<l;i++)
	{
		int x=i;
		int y=b2;
		while(str[x]==op[y]&&x<l&&y<11)
		{
			x++,y++;
		}
		dfs(x+1,y,num+1);
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		L=strlen(op);
		memset(str,'\0',sizeof(str));
		scanf("%s",str);
		flog=0;
		l=strlen(str);
		dfs(0,0,0);
		if(flog)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}