【2019杭电多校第五场1006=HDU6629】string matching(求后缀字符串和本串的最长公共前缀长度之和--扩展kmp)

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=6629

题目：

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string matching

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description

String matching is a common type of problem in computer science. One string matching problem is as following:

Given a string s[0…len−1], please calculate the length of the longest common prefix of s[i…len−1] and s[0…len−1] for each i>0.

I believe everyone can do it by brute force.
The pseudo code of the brute force approach is as the following:

We are wondering, for any given string, what is the number of compare operations invoked if we use the above algorithm. Please tell us the answer before we attempt to run this algorithm.

Input

The first line contains an integer T, denoting the number of test cases.
Each test case contains one string in a line consisting of printable ASCII characters except space.

* 1≤T≤30

* string length ≤106 for every string

Output

For each test, print an integer in one line indicating the number of compare operations invoked if we run the algorithm in the statement against the input string.

Sample Input

3

_Happy_New_Year_

ywwyww

zjczzzjczjczzzjc

Sample Output

17

7

32

解题思路：

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新算法get✅

扩展KMP裸题，不能用后缀数组，倍增算法nlogn超时，DC3算法常数过大，同样超时！！

扩展kmp讲解参见https://www.jianshu.com/p/107e47994d49

 

ac代码：

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#include 
using namespace std;
typedef long long ll;
const int maxn=1000010; //字符串长度最大值
int nxt[maxn];
char s[maxn];
ll  solve(char str[])
{
    ll ans = 0;
    int i=0,j,po,len=strlen(str);
    nxt[0]=len; //初始化nxt[0]
    while(str[i]==str[i+1] && i+1po+nxt[po],则要从头开始匹配
            while(i+j