hdu 5311 Hidden String (BestCoder 1st Anniversary ($))(深搜）

http://acm.hdu.edu.cn/showproblem.php?pid=5311

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1499    Accepted Submission(s): 534

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 ($)

题目大意：给两个字符串s，str[] = "anniversary";

将str分成三个部分问这三个部分能否在s中找到

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#define N 110

using namespace std;

int len;
char s[N], str[] = "anniversary";

bool DFS(int d, int l1, int l2)//d表示搜索的层次，l1表示从s的l1位置处开始，l2表示str从l2位置处开始，二者进行比较
{
    int i, a, b;
    if(l2 == )
        return true;//搜到str的完整序列，且之前搜索的层次不大于3
    if(d > )
        return false;//搜索的层次大于3则返回false
    for(i = l1 ; i < len ; i++)
    {
        a = i;//表示从s的a位置处开始
        b = l2;//表示从str的b位置处开始
        if(s[i] == str[l2])
        {
            while(s[a] == str[b] && a < len && b < )
            {
                a++;
                b++;
            }
            if(DFS(d + , a, b))
                return true;
        }
    }
    return false;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s", s);
        len = strlen(s);
        if(DFS(, , ))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return ;
}