写在前面
仅为自用，不做推广
一起来看猫片吧！
一篇不错的博客，然而我闷了一下午还是不会，看了看书算是搞懂了
博客里面各种性质讲的非常详细，有空可以回看一下
核心的两段代码
nxt数组预处理：
我这里使用pre表示nxt数组，用go表示要匹配的串
void init(){//预处理pre数组
int len = strlen(go + 1);
int j = 0;
for(int i = 1; i < len; ++i){
while(j > 0 && go[i + 1] != go[j + 1]) j = pre[j];
if(go[i + 1] == go[j + 1]) ++j;
pre[i + 1] = j;
}
}
原字符串的匹配：
for(int i = 0; i < len1; ++i){
while(j > 0 && s[i + 1] != go[j + 1]) j = pre[j];
if(s[i + 1] == go[j + 1]) ++j;
// cout<<"i:"<<i<<" "<<j<<endl;
if(j == len2){//如果匹配完成
cnt++;
j = 0;
}
}
例题
剪花布条
直接KMP匹配即可，匹配成功将匹配串的指针置为0
Radio Transmission
一个结论题，答案为 (n - nxt[n])，好像与nxt数组本身的性质有关
OKR-Periods of Words
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define orz cout<<"lkp AK IOI!"<<endl
using namespace std;
const int MAXN = 1e6+6;
const int INF = 1;
const int mod = 1;
int n;
LL ans = 0;
char s[MAXN];
int pre[MAXN];
int read(){
int s = 0, f = 0;
char ch = getchar();
while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();
while(isdigit(ch)) s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return f ? -s : s;
}
void init(){
int j = 0;
for(int i = 1; i <= n; ++i){
while(j > 0 && s[i + 1] != s[j + 1]) j = pre[j];
if(s[i + 1] == s[j + 1]) ++j;
pre[i + 1] = j;
}
}
int main()
{
n = read();
cin >> (s + 1);
init();
for(int i = 1; i <= n; ++i){
int j = i;
while(pre[j]) j = pre[j];
if(pre[i]) pre[i] = j;
ans += (i - j);
}
printf("%lld", ans);
return 0;
}
似乎在梦中见过的样子
看这位大佬的题解
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define orz cout<<"lkp AK IOI!"<<endl
using namespace std;
const int MAXN = 2e4+6;
const int INF = 1;
const int mod = 1;
int n, k, cnt = 0;
char s[MAXN];
int pre[MAXN];
int read(){
int s = 0, f = 0;
char ch = getchar();
while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();
while(isdigit(ch)) s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return f ? -s : s;
}
void Kmp(int l){
int j = l - 1;
pre[l] = pre[l - 1] = j;
for(int i = l; i < n; ++i){
while(j > l - 1 && s[j + 1] != s[i + 1]) j = pre[j];
if(s[j + 1] == s[i + 1]) j++;
pre[i + 1] = j;
}
for(int i = l; i < n; ++i){
j = pre[i + 1];
while(j > l - 1 && l + 2 * (j - l + 1) > i + 1) j = pre[j];
if(j - l + 1 >= k) cnt++;
}
}
int main()
{
cin >> (s + 1);
k = read();
n = strlen(s + 1);
for(int i = 1; i <= n; ++i) Kmp(i);
printf("%d", cnt);
return 0;
}
Censoring
主要思路是开一个栈，来储存还未被消去的字符串
如果一个串匹配完成，从弹出相应的串
在入栈是顺便记录入栈字符的失陪位置，匹配完一个串后可以直接从栈顶所对字符的失陪位置开始匹配
从前到后跑一遍即可，复杂度 (O(n))
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define orz cout<<"lkp AK IOI!"<<endl
using namespace std;
const int MAXN = 1e6+6;
const int INF = 1;
const int mod = 1;
char s[MAXN], t[MAXN];
int lens, lent;
int pre[MAXN], f[MAXN];
int stc[MAXN], sc = 0;
int read(){
int s = 0, f = 0;
char ch = getchar();
while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();
while(isdigit(ch)) s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return f ? -s : s;
}
void init(){
int j = 0;
for(int i = 1; i <= lent; ++i){
while(j && t[i + 1] != t[j + 1]) j = pre[j];
if(t[i + 1] == t[j + 1]) ++j;
pre[i + 1] = j;
}
}
int main()
{
cin >> (s + 1);
cin >> (t + 1);
lens = strlen(s + 1);
lent = strlen(t + 1);
init();
for(int i = 0, j = 0; i < lens; ++i){
while(j && s[i + 1] != t[j + 1]) j = pre[j];
if(s[i + 1] == t[j + 1]) ++j;
f[i + 1] = j;
stc[++sc] = i + 1;
if(j == lent){
sc -= lent, j = f[stc[sc]];
}
}
for(int i = 1; i <= sc; i++){
printf("%c", s[stc[i]]);
}
return 0;
}