HDOJ5311 深搜简单题，几个小的剪枝_hdoj搜索剪枝例题-优快云博客

本文链接：https://blog.youkuaiyun.com/computer_user/article/details/77073567

, 本文介绍了一个编程问题，任务是在给定字符串中找到三个不重叠的子字符串，使其拼接后形成anniversary。文章提供了问题的具体描述、输入输出样例及示例代码。

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Hidden String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2018 Accepted Submission(s): 766

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

Source

BestCoder 1st Anniversary ($)

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代码如下：

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
using namespace std;

string s,word="anniversary";
bool flag;
int lens,lenw;

void dfs(int num, int index,int word_index){
   if (flag) return;
   if (num<=3 && word_index>=lenw) { 
      flag=1;
      return;
   }
   if (index>=lens) return;
   if (word_index>=lenw) return;
   if (num>3) return;

   int i=index;
   int k=word_index;
   while (i<lens){
      if (s[i]==word[k]){
          int tmp=i;
         while (s[tmp]==word[k] && tmp<lens && k<lenw) {
              tmp++; k++;
         };
         dfs(num+1,tmp,k);
         if (flag) return;
         k=word_index;
      }
      i++;
   }
}

int main(){
    std::ios::sync_with_stdio(false);
    lenw=word.length(); 
    int t;
    cin>>t;
    while (t--){
        cin>>s;
        flag=0;
        lens=s.length();
        dfs(0,0,0);
        if (flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}