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

2
annivddfdersewwefary
nniversarya

输出样例

YES
NO

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;
	else
		return 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;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}

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