POJ --- 1200 Crazy Search

Crazy Search

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21585		Accepted: 6101

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

5

思路：利用Karp-Rabin算法的思想，对每个子串进行Hash，如果Hash值相等则认为这两个子串是相同的（事实上还需要做进一步检查），Karp-Rabin算法的Hash函数有多种形式，但思想都是把字符串映射成一个数字。本题hash函数是把字串转化为NC进制的数（实际上程序中计算结果已经被转换为10进制，因为NC进制数不同转化为10进制数自然不同，所以不影响判断结果），数组开到了1.6×10^7（我试了一下1.2×10^7也能AC）,实际上这也是不严谨的，因为我们不能保证hash之后的数值在这个范围内，比如N=NC=35,程序就有Bug了，但是这题后台数据可能没这么给。在实际运用中是需要取模的，而且即使hash值相等也需要进一步比对。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
bool hash[16000005];
int is_have[300];
char str[1000005];
int main(){
    int n, nc;
    /* freopen("in.c", "r", stdin); */
    while(~scanf("%d%d", &n, &nc)){
        memset(str, 0, sizeof(str));
        memset(hash, 0, sizeof(hash));
        memset(is_have, 0, sizeof(is_have));
        scanf("%s", str);
        int len = strlen(str);
        int k = 0, ans = 0;
        for(int i = 0;i < len;i ++) is_have[str[i]] = 1;
        for(int i = 0;i < 256;i ++)
            if(is_have[i]) is_have[i] = k++;
        for(int i = 0;i <= len - n;i ++){
            int key = 0;
            for(int j = i;j < i + n;j ++) key = key*nc + is_have[str[j]];
            if(!hash[key]) ans ++, hash[key] = 1;
        }
        printf("%d
", ans);
    }
    return 0;
}

 下面附上一般情况的实现代码(均来自其他网友）：

1. 原文链接：http://www.xefan.com/archives/83853.html

#include <stdio.h>
#include <math.h>
int mod = 0x7fffffff;
const int d = 128;
int rabin_karp(char *T, char *P, int n, int m)
{
    if (n < m) return -2;
    int h = pow(d, m-1);
    int p = 0;
    int t = 0;
    int i, j;
    for (i=0; i<m; i++) {
        p = (d*p + P[i]) % mod;
        t = (d*t + T[i]) % mod;
    }
    for (j=0; j<=n-m; j++) {
        if (p == t) {
            return j;
        }
        if (j < n-m) {
            t = (d*(t - h*T[j]) + T[j+m]) % mod;
        }
    }
    return -1;
}
 
int main(int argc, char *argv[])
{
    char t[] = "BBC ABCDAB ABCDABCDABDE";
    char p[] = "ABCDABD";
    int len1 = sizeof(t) - 1;
    int len2 = sizeof(p) - 1;
    int index = rabin_karp(t, p, len1, len2);
    printf("index: %d
", index);
    return 0;
}

 2.

原文链接：http://blog.csdn.net/onezeros/article/details/5531354

//Karp-Rabin algorithm,a simple edition
int karp_rabin_search(const char* text,const int text_len,const char* pattern,const int pattern_len)
{
    int hash_text=0;
    int hash_pattern=0;
    int i;
    
    //rehash constant:2^(pattern_len-1)
    int hash_const=1;
    /*for (i=1;i<pattern_len;i++){
        hash_const<<=1;
    }*/
    hash_const<<=pattern_len-1;

    //preprocessing
    //hashing
    for (i=0;i<pattern_len;++i){
        hash_pattern=(hash_pattern<<1)+pattern[i];
        hash_text=(hash_text<<1)+text[i];
    }

    //searching
    for (i=0;i<=text_len-pattern_len;++i){
        if (hash_pattern==hash_text&&memcmp(text+i,pattern,pattern_len)==0){
            return i;
        }else{
            //rehash
            hash_text=((hash_text-text[i]*hash_const)<<1)+text[i+pattern_len];
        }
    }
    return -1;
}