poj1200（字符串Hash）

最新推荐文章于 2019-10-30 17:24:56 发布

原创最新推荐文章于 2019-10-30 17:24:56 发布 · 289 阅读

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本文介绍了一个计算机科学中的字符串处理问题——疯狂搜索挑战。任务要求找出指定大小的所有不同子串数量，并提供了一种有效的算法实现，利用散列表记录子串出现情况，以避免重复计数。

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Crazy Search

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30686		Accepted: 8503

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

#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<list>
#include<set>
#include<iomanip>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cassert>
#include<sstream>
#include<algorithm>
using namespace std;
const int MAXN = 2e7;
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
typedef long long ll;

bool hash[MAXN];
char w[MAXN];
int id[500]; //字符最多为256

int main()
{
	int nc,n,i,j;
	while(~scanf("%d%d",&n,&nc))
	{
		memset(hash,0,sizeof(hash));
		memset(id,-1,sizeof(id));
		int cnt=0;
		scanf("%s",w);
		int len=strlen(w);
		for(i=0; i<len&&cnt<nc; i++)//nc种字符转化为0，2，...，nc-1
		{
			if(id[w[i]]!=-1) continue;
			id[w[i]]=cnt++;
		}
		int ans=0;
		for(i=0; i<len-n+1; i++)//枚举长度为n的子串
		{
			int s=0;
			for(j=i; j<i+n; j++)
			{
				s=s*nc+id[w[j]];//转化为相应的nc进制
			}
			if(hash[s]) continue;//如果已存在，到下一个字串
			ans++;//如果之前未存在，ans++
			hash[s]=true;
		}
		printf("%d\n",ans);
	}
	return 0;
}