HDU 5311：Hidden String

Hidden String

 

 Accepts: 437

 

 Submissions: 2174

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 262144/262144 K (Java/Others)

问题描述

今天是BestCoder一周年纪念日. 比赛管理员Soda有一个长度为n的字符串s. 他想要知道能否找到s的三个互不相交的子串s[l1..r1], s[l2..r2], s[l3..r3]满足下列条件:  1. 1≤l1≤r1<l2≤r2<l3≤r3≤n  2. s[l1..r1], s[l2..r2], s[l3..r3]依次连接之后得到字符串"anniversary".

输入描述

输入有多组数据. 第一行有一个整数T (1≤T≤100), 表示测试数据组数. 然后对于每组数据:一行包含一个仅含小写字母的字符串s (1≤|s|≤100).

输出描述

对于每组数据, 如果Soda可以找到这样三个子串, 输出"YES", 否则输出"NO".

输入样例

2annivddfdersewwefarynniversarya

输出样例

YESNO

这题想要总结的地方就是理解一下题意吧，分成三段。。。(单独的annniversary要输出YES)

改了很久也很low的代码，也不优化了，纪念一下吧

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <cstring>#pragma warning(disable:4996)using namespace std;vector<int> wei_z[30];vector<int> weizhi;char test[200];int len,i,flag;void init(){for(i=0;i<len;i++){wei_z[test[i]-'a'].push_back(i);}}bool pend(){int result=0,temp=weizhi[0];for(i=1;i<weizhi.size();i++){if(temp!=weizhi[i]-1)result++;temp=weizhi[i];}if(result<=2)return true;elsereturn false;}void solve(){int i1,i2,i3,i4,i5,i6,i7,i8,i9,i10,i11;for(i1=0;i1<wei_z['a'-'a'].size();i1++){for(i2=0;i2<wei_z['n'-'a'].size();i2++){if(wei_z['n'-'a'][i2]<=wei_z['a'-'a'][i1])continue;for(i3=0;i3<wei_z['n'-'a'].size();i3++){if(wei_z['n'-'a'][i3]<=wei_z['n'-'a'][i2])continue;for(i4=0;i4<wei_z['i'-'a'].size();i4++){if(wei_z['i'-'a'][i4]<=wei_z['n'-'a'][i3])continue;for(i5=0;i5<wei_z['v'-'a'].size();i5++){if(wei_z['v'-'a'][i5]<=wei_z['i'-'a'][i4])continue;for(i6=0;i6<wei_z['e'-'a'].size();i6++){if(wei_z['e'-'a'][i6]<=wei_z['v'-'a'][i5])continue;for(i7=0;i7<wei_z['r'-'a'].size();i7++){if(wei_z['r'-'a'][i7]<=wei_z['e'-'a'][i6])continue;for(i8=0;i8<wei_z['s'-'a'].size();i8++){if(wei_z['s'-'a'][i8]<=wei_z['r'-'a'][i7])continue;for(i9=0;i9<wei_z['a'-'a'].size();i9++){if(wei_z['a'-'a'][i9]<=wei_z['s'-'a'][i8])continue;for(i10=0;i10<wei_z['r'-'a'].size();i10++){if(wei_z['r'-'a'][i10]<=wei_z['a'-'a'][i9])continue;for(i11=0;i11<wei_z['y'-'a'].size();i11++){if(wei_z['y'-'a'][i11]<=wei_z['r'-'a'][i10])continue;weizhi.push_back(wei_z['a'-'a'][i1]);weizhi.push_back(wei_z['n'-'a'][i2]);weizhi.push_back(wei_z['n'-'a'][i3]);weizhi.push_back(wei_z['i'-'a'][i4]);weizhi.push_back(wei_z['v'-'a'][i5]);weizhi.push_back(wei_z['e'-'a'][i6]);weizhi.push_back(wei_z['r'-'a'][i7]);weizhi.push_back(wei_z['s'-'a'][i8]);weizhi.push_back(wei_z['a'-'a'][i9]);weizhi.push_back(wei_z['r'-'a'][i10]);weizhi.push_back(wei_z['y'-'a'][i11]);if(pend()==true){flag=1;return;}weizhi.clear();}}}}}}}}}}}}int main(){int Test1;cin>>Test1;while(Test1--){cin>>test;for(i=0;i<=29;i++)wei_z[i].clear();weizhi.clear();flag=0;len=strlen(test);init();solve();if(flag)cout<<"YES"<<endl;elsecout<<"NO"<<endl;}return 0;}