HDU - 3336 Count the string（KMP+DP）

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本文介绍了一个字符串处理问题：计算一个给定字符串的所有非空前缀作为模式串在该字符串中匹配的总次数，并提供了一种使用KMP算法的解决方案。

Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11320 Accepted Submission(s): 5276

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

Sample Output

Author

foreverlin@HNU

Source

HDOJ Monthly Contest – 2010.03.06

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题意：给出主串，以主串的所有前缀为模式串问所有模式串匹配次数之和；

#include <bits/stdc++.h>
using namespace std;

const int N = 200000 + 10, Mod = 10007;
int t, n;
char a[N];
int nex[N], dp[N];

void get_nex(){
    int i = 0, j = -1;
    nex[0] = -1;
    while(i < n){
        if(j == -1 || a[i] == a[j]){
            i++, j++;
            nex[i] = j;
        }
        else j = nex[j];
    }
}

int main(){
    scanf("%d", &t);
    while(t--){
        memset(dp, 0, sizeof dp);
        scanf("%d", &n);
        scanf("%s", a);
        get_nex();
        int ans = 0;
        for(int i = 1; i <= n; i++){
            dp[i] = dp[nex[i]]+1;
            ans += dp[i];
            ans %= Mod;
        }
        printf("%d\n", ans);
    }
}