杭电6629string matching_string matching is a common type of problem in com-优快云博客

本文探讨了字符串匹配中的一项常见挑战：计算任意子串与其原串的最长公共前缀长度。通过一种改进的KMP算法实现，文章详细介绍了算法的实现原理与步骤，包括next数组的构建和比较操作次数的计算，旨在优化字符串匹配效率。

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题目描述

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.

输入

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

输出

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.

样例输入

复制样例数据

3
_Happy_New_Year_
ywwyww
zjczzzjczjczzzjc

样例输出

17
7
32

扩展kmp板题，只用到next数组就可以

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll mod=1e9+7;
const int modp=998244353;
const int maxn=1e6+50;
const double eps=1e-6;
#define lowbit(x)  x&(-x)
#define INF 0x3f3f3f3f
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
int dcmp(double x)
{
    if(fabs(x)<eps)return 0;
    return (x>0)?1:-1;
}
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
ll qmod(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%mod;
        }
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
char str[maxn];
int nex[maxn];
ll getNext(char str[])
{
    int i=0,j,po,len=strlen(str);
    ll ans=0;
    nex[0]=len; //初始化next[0]
    while(str[i]==str[i+1] && i+1<len) i++; nex[1]=i; //计算next[1]
    ans+=nex[1];
    if(nex[1]!=len-1)ans++;
    po=1; //初始化po的位置
    for(i=2;i<len;i++)
    {
        if(nex[i-po]+i < nex[po]+po) //第一种情况，可以直接得到next[i]的值
            nex[i]=nex[i-po];
        else //第二种情况，要继续匹配才能得到next[i]的值
        {
            j = nex[po]+po-i;
            if(j<0) j=0; //如果i>po+next[po],则要从头开始匹配
            while(i+j<len && str[j]==str[j+i]) j++; nex[i]=j;
            po=i; //更新po的位置
        }
        ans+=nex[i];
        if(nex[i]!=len-i)ans++;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",str);
        printf("%lld\n",getNext(str));
    }
    return 0;
}