hdu1381Crazy Search hash map 模板题

Crazy Search

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3087    Accepted Submission(s): 1154

Problem Description

Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in the text. As you soon will discover, you really need the help of a computer and a good algorithm to solve such a puzzle.

Your task is to write a program that given the size, N, of the substring, the number of different characters that may occur in the text, NC, and the text itself, determines the number of different substrings of size N that appear in the text.

As an example, consider N=3, NC=4 and the text "daababac". The different substrings of size 3 that can be found in this text are: "daa", "aab", "aba", "bab", "bac". Therefore, the answer should be 5. 

 

Input

The first line of input consists of two numbers, N and NC, separated by exactly one space. This is followed by the text where the search takes place. You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions.

 

Output

The program should output just an integer corresponding to the number of different substrings of size N found in the given text.
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

Sample Input

13 4daababac

 

Sample Output

5

 

Recommend

第一种 hash哈希表 有点复杂相较于第二种
#include<stdio.h>
#include<string.h>
using namespace std;
#define SIZE 16000005
#define mod  16000005 
using namespace std;
 
int n,nc;
char str[SIZE];
int hash[30];
bool vis[SIZE];

int main()
{
	int T;
	scanf("%d",&T);
	for(int t=1;t<=T;t++)
	{
		scanf("%d%d",&n,&nc);
		scanf("%s",str+1);
		memset(hash,0,sizeof(hash));
		memset(vis,0,sizeof(vis));
		int len=(int)strlen(str+1);
		int val=0;
		for(int i=1;i<=len;i++)//对每个不同的字符对应到不同的权值 
		{
			if(!hash[str[i]-'a'])
			hash[str[i]-'a']=++val;
			if(val==nc)
			break;
		}
		int ans=0;
		for(int i=1;i<=len-n+1;i++)//枚举子串 注意循环变量i和j的范围 
		{
			int sum=0;
			for(int j=i;j<=i+n-1;j++)
			sum=sum*nc+hash[str[j]-'a'];
			
			sum%=mod;
			if(!vis[sum])
			{
				vis[sum]=true;
				ans++;
			}			
		}
		printf("%d\n",ans);
		if(t!=T)
		printf("\n");
	}
	return 0;
}

第二种 用map映射
#include<iostream>
#include<string>
#include<map>
using namespace std;
int main()
{
	int t,n,nc;
	cin>>t;
	while(t--)
	{
		string s;
		map<string,int> smap;//map映射 
		cin>>n>>nc>>s;
		int k=0;
		int len=s.length();
		for(int i=0;i<=len-n;i++)
		{
			string ss=s.substr(i,n);//提取子串 
			if(smap[ss]==0)
			{
				k++;
				smap[ss]=1;
			}
		}
		cout<<k<<endl;
		if(t>1)
		cout<<endl;
	}
	
	return 0;
}