Hidden String(递归)

最新推荐文章于 2021-02-26 04:17:03 发布

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本文探讨了在给定字符串中查找三个不重叠子串的问题，这些子串的拼接形成目标字符串anniversary。通过代码实现，解决了如何在多测试案例下判断是否存在符合条件的子串。

Hidden String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 595 Accepted Submission(s): 225

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

(1≤T≤100), indicating the number of test cases. For each test case:

There's a line containing a string

(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

#include <stdio.h>
#include <string.h>

using namespace std;

char str[110];
char s[] = "anniversary";
int len1 = strlen(s);
int len;
bool flag;

bool DFS(int i,int j,int sum)
{
    if(sum > 3)
        return false;
    int k;
    for(k=0;i+k<len&&j+k<len1;k++)
    {
        if(str[i+k] != s[j+k])
        {
            for(int ans = i+k+1;ans<len;ans++)
            {
                if(str[ans] == s[j+k])
                {
                    flag = DFS(ans,j+k,sum+1);
                    if(flag)return true;
                }
            }
            return false;
        }
    }
    if(j+k == len1 && sum<=3)
        return true;
    return false;
}

int main()
{
    freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        len = strlen(str);
        if(len < len1)
        {
            printf("NO\n");
            continue;
        }
        flag = false;
        for(int i=0;i<len;i++)
        {
            if(str[i] == s[0])
            {
                flag = DFS(i,0,1);
                if(flag)break;
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}