【KMP+DP】Count the string

KMP算法的综合练习

DP很久没写搞了半天才明白。本题结合Next[]的意义以及动态规划考察对KMP算法的掌握。

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

Author

foreverlin@HNU

Source

HDOJ Monthly Contest – 2010.03.06

 

#include<iostream>  //KMP+DP
#include<memory.h>
using namespace std;
char s[200005];
int Next[200005],DP[200005];    //DP[i]表示子串s[0~i]共含有以s[i]为结尾的前缀的数目
int l;

void GetNext(){
    int i=0,j=-1;
    Next[0]=-1;
    while(i<l){
        if(j==-1||s[i]==s[j]){
            i++;
            j++;
            Next[i]=j;
        }
        else
            j=Next[j];
    }
}

int main()
{
    int n,k,num;
    cin>>n;
    while(n--){
        cin>>l>>s;
        GetNext();
        num=0;
        memset(DP,0,sizeof(DP));
        for(k=1;k<=l;k++){
            DP[k]=DP[Next[k]]+1;    //s[i]结尾的前缀数就是自己本身加上以s[Next[i]]结尾的前缀数
            num=(num+DP[k])%10007;
        }
        cout<<num<<endl;
    }
    return 0;
}