Hdu 3336 Count the string(求给定字符串含前缀的数量)

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7470    Accepted Submission(s): 3457

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

   
   
   
   

    
    
    
    6
   
   
   
   

题意:

求给定字符串含前缀的数量 (1 <= n <= 200000)

Abab,前缀为a,ab,aba,abab

abab中共有六个子串是前缀a a ab ab aba abab

所以答案为6

 

方法一:

Kmp+dp

Dp[i]=dp[Next[i]]+1,dp[i]表示以i为结尾包含的字符串的数量。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1100000;
int n,m;
const int MOD=10007;
char pattern[maxn];
int Next[maxn],ans,dp[maxn];

bool solve(int x1,int y1,int x2,int y2){
    for(int i=x1,j=x2;i<=y1,j<=y2;i++,j++)
        if(pattern[i]!=pattern[j])
            return false;
    return true;
}

void get_next(){
    int i=0,j=-1;
    Next[0]=-1,dp[0]=0;
    while(i<m){
        if(j==-1||pattern[j]==pattern[i]){
            j++,i++;
            Next[i]=j;
            dp[i]=dp[Next[i]]+1;
            ans+=dp[i];
        }
        else
            j=Next[j];
    }
}


int main(){
    int _;
    scanf("%d",&_);
    while(_--){
        scanf("%d",&m);
        scanf("%s",pattern);
        ans=0;
        get_next();
        printf("%d\n",ans%MOD);
    }
    return 0;
}

方法二:

EKMP

Next[i]表示以i为开头的字符串与与0为开头的字符串的最大公共前缀。

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=210000;
char x[maxn];
int Next[maxn];

void pre_EKMP(int m){
    Next[0]=m;
    int j=0;
    while(j+1<m&&x[j]==x[j+1])
        j++;
    Next[1]=j;
    int k=1;
    for(int i=2;i<m;i++){
        int p=Next[k]+k-1;
        int L=Next[i-k];
        if(i+L<p+1)
            Next[i]=L;
        else{
            j=max(0,p-i+1);
            while(i+j<m&&x[i+j]==x[j])
                j++;
            Next[i]=j;
            k=i;
        }
    }
}

int main(){
    int _,n;
    scanf("%d",&_);
    while(_--){
        scanf("%d",&n);
        scanf("%s",x);
        pre_EKMP(n);
        int ans=0;
        for(int i=0;i<n;i++)
            ans+=Next[i];
        printf("%d\n",ans%10007);
    }
    return 0;
}