POJ 1200字符串hash

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本文介绍了一个算法挑战，即在一个给定的字符串中寻找所有不同长度为N的子串数量，并通过哈希方法解决了这一问题。

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Crazy Search

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30412		Accepted: 8408

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.

Sample Input

3 4
daababac

Sample Output

Hint

Huge input,scanf is recommended.

Source

Southwestern Europe 2002

题意：给出一个字符串，其中包含NC种字符，求这个字符串种长度为N的子串有多少种（不重复的数量）。

题解：将字符串种所有的长度为N的字串进行hash处理，hash为NC进制的数，最后统计个数即可。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define N 16000010

using namespace std;

char s[N];
bool has[N]; //hash数组
int id[1010]; //记录每个字符转化成的数字

int main()
{
int n,nc;
while(~scanf("%d %d",&n,&nc))
{
scanf("%s",s);
int len=strlen(s);
int cnt=0;
memset(has,false,sizeof(has)); //初始化
memset(id,-1,sizeof(id));
for(int i=1;i<len&&cnt<nc;i++)
{
if(id[s[i]]!=-1) continue;
id[s[i]]=cnt++;
}
int ans=0;
for(int i=0;i<len-n+1;i++)
{
int tmp=0;
for(int j=i;j<i+n;j++)
{
tmp=tmp*nc+id[s[j]]; //将字串转化为NC进制的数字
}
if(has[tmp]) continue;
ans++; //ans统计个数
has[tmp]=true;
}
printf("%d\n",ans);
//cout << ans << endl;
}
return 0;
}