hdu 6629 string matching【2019 Multi-University Training Contest 5】【扩展kmp】

string matching

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 630    Accepted Submission(s): 240

 

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

题目链接：

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

题意：

按照题目所给的代码计算字符串比对次数，实际就是比较母串【0-len】与母串的子串【1-len】。

题解：

扩展kmp板子题，一开始预处理next数组时，将每一位的next求和即可，要记得不想等要+1，因为比较了一次不想等

据说数位dp也能过（这也太硬核了），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！！

代码：

#include
using namespace std;
#define ll long long
const int maxn = 1e6+5;

ll t,n,ans;
ll nxt[maxn];
char ch[maxn];

void get_next(char *str) {
	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>t;
	while(t--) {
		ans = 0;
		scanf("%s",ch);
		get_next(ch);
		cout<