hdoj 5311 Hidden String 【dfs】

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1730    Accepted Submission(s): 606

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a strings 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 integerT(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

 

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char a[12]="anniversary";char s[110];int l;int f;void dfs(int p1,int p2,int v){if(v>3)return ;if(p2>=11)    {    f=1;    return ;    }    for(int i=p1;i<l;i++)    {    int x=i,y=p2;    while(s[x]==a[y])    {    x++;    y++;    }    dfs(x+1,y,v+1);    }}int main(){int T;scanf("%d",&T);while(T--){f=0;scanf("%s",s);    l=strlen(s);    dfs(0,0,0);    if(f)    printf("YES\n");    else    printf("NO\n");}return 0;}