[HDU] 5311 Hidden String [DFS]_you are given string s s and t t consisting of low-优快云博客

本文介绍了一个字符串匹配的问题，即在一串给定的字符中寻找三个不重叠的子串，使得它们连接起来能构成特定字符串“anniversary”。文章提供了一种递归深度优先搜索的解决方案，并通过多个测试案例验证了其正确性。

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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 $s[l_1..r_1]$ , $s[l_2..r_2]$ , $s[l_3..r_3]$ that:

$1 \le l_1 \le r_1 < l_2 \le r_2 < l_3 \le r_3 \le n$
The concatenation of $s[l_1..r_1]$ , $s[l_2..r_2]$ , $s[l_3..r_3]$ is “anniversary”.

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

There’s a line containing a string $s$ $(1 \le |s| \le 100)$ 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).

题解

这么简单的题，粗心的我WA了几发QAQ

代码如下

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

char S[]="anniversary";
char s[120];
int L=strlen(S);
int len;

bool dfs(int cnt,int s_i,int S_i){
    if(S_i==L&&cnt>=0) return true;
    if(s_i==len||cnt<0) return false;
    for(int i=s_i;i<len;i++){
        if(s[i]==S[S_i])
            if(dfs(cnt+(i>s_i?-1:0),i+1,S_i+1))
                return true;
    }
    return false;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s+1);
        len=strlen(s+1)+1;
        puts(dfs(3,0,0)?"YES":"NO");
    }
    return 0;
}