HDU 3336 Count the string （扩展KMP）

最新推荐文章于 2020-07-06 19:54:07 发布

LSD20164388

最新推荐文章于 2020-07-06 19:54:07 发布

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分类专栏：字符串处理：KMP 文章标签： ACM

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本文深入探讨了扩展KMP算法的应用，特别是在字符串匹配问题中，如何高效计算所有前缀在字符串中的出现次数总和，并通过实例说明了算法的实现过程。

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Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15427 Accepted Submission(s): 7040

Problem Description

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

Input

The first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.

Output

For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.

Sample Input

1 4 abab

Sample Output

题意：

给你一个长度为n(n<=2e5)的串，求其每个前缀在字符串中的数量之和%10007。

思路：

标准的扩展KMP。答案就是每个next数组的和。（扩展KMP不懂的小伙伴可以去百度~）

代码：

#include<iostream>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<queue>
#include<vector>
#include<map>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1000010;
const int K=1000005;
int nt[K],extand[K],n;
char S[K],T[K];
void Getnext(char *T,int *next)
{
    int len=strlen(T),a=0;
    next[0]=len;
    while(a<len-1 && T[a]==T[a+1]) a++;
    next[1]=a;
    a=1;
    for(int k=2; k<len; k++)
    {
        int p=a+next[a]-1,L=next[k-a];
        if( (k-1)+L >= p)
        {
            int j = (p-k+1)>0 ? (p-k+1) : 0;
            while(k+j<len && T[k+j]==T[j]) j++;
            next[k]=j;
            a=k;
        }
        else
            next[k]=L;
    }
}
void GetExtand(char *S,char *T,int *next)
{
    Getnext(T,next);
    int slen=strlen(S),tlen=strlen(T),a=0;
    int MinLen = slen < tlen ? slen : tlen;
    while(a<MinLen && S[a]==T[a]) a++;
    extand[0]=a;
    a=0;
    for(int k=1; k<slen; k++)
    {
        int p=a+extand[a]-1, L=next[k-a];
        if( (k-1)+L >= p)
        {
            int j= (p-k+1) > 0 ? (p-k+1) : 0;
            while(k+j<slen && j<tlen && S[k+j]==T[j]) j++;
            extand[k]=j;
            a=k;
        }
        else
            extand[k]=L;
    }
}
int main()
{
    int TT,cas=1;
    scanf("%d",&TT);
    while(TT--)
    {
        scanf("%d",&n);
        scanf("%s",T);
        Getnext(T,nt);
        int ans=0;
        for(int i=0; i<n; i++)
        ans=(ans+nt[i])%10007;
        printf("%d\n",ans);
    }
    return 0;
}