HDOJ 5311-Hidden String【深搜】

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

2annivddfdersewwefarynniversarya

 

Sample Output

YESNO

 

解题思路

因为这个题中长度比较小可以暴力搞一下，利用深搜写在判断里判断是否满足段数大于三，然后继续深搜。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int len,len1;char map[100000];char moxing[20]="anniversary";bool f(int x,int y,int ci){int i;if(ci<=3&&y>=len1){return true;}else if(ci>3){return false;}for(i=x;i<len;i++){int wc,wcnm;wc=i,wcnm=y;while(wc<len&&wcnm<len1&&map[wc]==moxing[wcnm]){wc++;wcnm++;}if(wc>i&&f(wc,wcnm,ci+1)){return true;}}return false;}int main(){int t;scanf("%d",&t);while(t--){scanf("%s",map);len=strlen(map);len1=strlen(moxing);bool cnm=f(0,0,0);if(cnm==true)printf("YES");elseprintf("NO");printf("\n");}return 0;}