kmp算法

Posted 2021-02-11 downrainsun

https://blog.csdn.net/dl962454/article/details/79910744

https://blog.csdn.net/qq_40938077/article/details/80460853

https://blog.csdn.net/liujiuxiaoshitou/article/details/70232219

 

https://vjudge.net/contest/240809#problem/A习题链接

Number Sequence

 HDU - 1711 

//这道题的next数组是由优化的next数组写的，next数组的含义不是指最长前缀与后缀的长度，因为加了递归优化，值一般就是-1与0.
//求最开始的匹配位置。（题目意思）
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
const int maxm = 1e6 + 5;




int a[maxm], b[maxm], nex[maxm];
int t, n, m, x;

void getnext() {
nex[0] = -1;
int k = -1;
int j = 0 ;
while(j < m - 1) {
    if(k == -1 || b[j] == b[k]) {
        j++, k++;
        if(b[j] != b[k]) nex[j] = k;
        else nex[j] = nex[k];
    }
    else k = nex[k];

}
}

int kmps() {
int i = 0, j = 0;
getnext();
while(i < n && j < m) {
    if(j == -1 || a[i] == b[j]) {
        i++, j++;
    }
    else j = nex[j];
//    printf("cbuisdc
");
}
if(j == m) return i - j + 1;
return -1;
}


int main() {
scanf("%d", &t);
while(t--) {
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    for(int i = 0; i < m; i++) {
        scanf("%d", &b[i]);
    }
    if(kmps() == -1) printf("-1
");
    else printf("%d
", kmps());
}
return 0;
}

Period

 HDU - 1358 

从这一题把next数组输出可知，next数组值该字符串的最长相同前缀与后缀，而且是包含了本身的。（如果这么说的话，相当于把字符串下标从1开始）

//这个是没加递归优化的next数组，指的是最长前缀与后缀的值。
//题目意思是指在i之前的字符串是不是有循环节的，有的话就输出循环节的长度与循环节的个数。
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

const int maxm = 1e6 + 5;

int n;
int nex[maxm];
void getnext(char ch[], int m) {
memset(nex, 0, sizeof(nex));
nex[0] = -1;
int j = 0, k = -1;
while(j < m) {
    if(k == -1 || ch[j] == ch[k]) {
        j++;
        k++;
        nex[j] = k;
    }
    else k = nex[k];
}
//for(int i = 0; i <= m; i++) {
//    printf("%d 
", nex[i]);
//}
//printf("
");
}
char ch[maxm];
int main() {
int cnt = 0;
while(~scanf("%d", &n) && n) {
    scanf("%s", ch);
    getnext(ch, n);
    printf("Test case #%d
", ++cnt);
    for(int i = 2; i <= n; i++) {
        int j = i - nex[i];
        if(j != i) {
            if(i % j == 0) {
                printf("%d %d
", i, i / j);
            }
        }
    }
    printf("
");
}
return 0;
}

 Simpsons’ Hidden Talents

 HDU - 2594 

//这个nex数组是用的最长前缀与后缀，题目意思是指求两个字符串的最长相同的前缀与后缀。
//这个在kmp匹配那里有不同，重点看一下。还有一种做法是直接把两个字符串拼接起来，在中间加一个特殊字符，然后在用kmp去求最长前缀与后缀的长度。

/*
@Author: Top_Spirit
@Language: C++
*/
//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 5e4 + 10 ;
const int INF = 0x3f3f3f3f ;
const double PI = acos(-1.0) ;
const ull seed = 133 ;

string s1, s2 ;
int len1, len2 ;
int Next[Maxn] ;

void getNext(){
    Next[0] = -1 ;
    int j =0, k = -1 ;
    while (j < len1 ){
        if (k == -1 || s1[j] == s1[k]) Next[++j] = ++k ;
        else k = Next[k] ;
    }
}

void KMP (){
    int i = 0, j = 0 ;
    while (i < len2){
        if (j == -1 || s1[j] == s2[i]){
            i++ ;
            j++ ;
        }
        else j = Next[j] ;

    }
    if (j){
        for (int k = 0; k < j; k++) cout << s1[k] ;
        cout << " " << j << endl ;
    }
    else cout << 0 << endl ;
}

int main (){
    while (cin >> s1 >> s2 ){
        len1 = s1.size() ;
        len2 = s2.size() ;
        getNext() ;
        KMP() ;
    }
    return 0 ;
}