[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher(1) A-H

https://vjudge.net/contest/70325#overview

A::给你两个串a,b.若存在a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. 输出最小的k.

思路：kmp返回k的位置就好了

ACcode:

#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <iostream>using namespace std;int a[1000010];int b[10010],nextt[10010];void getNext(int b[],int len){    int m=len;    nextt[1]=0;    for(int k=0,q=2;q<=m;q++)    {        while(k>0 && b[k+1]!=b[q])        k=nextt[k];        if(b[k+1]==b[q])        k++;        nextt[q]=k;    }}int kmp(int a[],int b[],int LenA,int LenB){    int n=LenA,m=LenB;    getNext(b,m);    for(int k=0,q=1;q<=n;q++)    {        while(k>0 && b[k+1]!=a[q])            k=nextt[k];        if(b[k+1]==a[q])        k++;        if(k==m)            return q-m+1;    }    return -1;}int main(){    ios::sync_with_stdio(false);cin.tie(0);    int T,n,m;    cin>>T;    while(T--)    {       cin>>n>>m;       for(int i=1;i<=n;i++) cin>>a[i];       for(int i=1;i<=m;i++) cin>>b[i];       cout<<kmp(a,b,n,m)<<endl;    }}

B：两个串a,b .输出b在a中出现的次数

思路：kmp模版

ACcode:

#include <cstring>#include <cstdio>#define t_size 1000000#define p_size 10000char t[t_size+5];char p[p_size+5];int next[p_size+5];void prefix(char p[]){int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){while(k>0 && p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k;}}int kmp(char t[],char p[]){prefix(p);int n=strlen(t+1),m=strlen(p+1);int sum=0;for(int i=0,q=1;q<=n;q++){while(i>0 && p[i+1]!=t[q])i=next[i];if(p[i+1]==t[q])i++;if(i==m){sum++;i=next[i];}}return sum;}int main(void){int T;scanf("%d",&T);while(T--){scanf("%s",p+1);scanf("%s",t+1);int ans=kmp(t,p);printf("%d\n",ans);}return 0;} 

C：一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？

思路：可以用拓扑，但这里我使用KMP扩展。求出extend数组，匹配到的就加1，找到一个就跳过一个模式串的长度，继续找。

ACcode:

#include <cstdio>#include <cstring>#define size 1005char str[size];char mode[size];int next[size],extend[size]; void getNext(char mode[],int next[],int modeLen){int i,a,p;a=p=0;next[0]=modeLen;for(i=1;i<modeLen;i++){if(i>=p || i+next[i-a]>=p){if(i>=p) p=i;while(p<modeLen && mode[p]==mode[p-i]) ++p;next[i]=p-i;a=i;}else next[i]=next[i-a];}}void getExtend(char str[],int strLen,int extend[],char mode[],int modeLen){getNext(mode,next,modeLen);int a=0,p=0,sum=0;for(int i=0;i<strLen;i++){if(i>=p || i+next[i-a]>=p){if(i>=p) p=i;while(p<strLen && p-i <modeLen && str[p]==mode[p-i])p++;extend[i]=p-i,a=i;}else extend[i]=next[i-a];}}int main(){while(scanf("%s",str)==1 &&str[0]!='#'){scanf("%s",mode);int strLen=strlen(str),modeLen=strlen(mode);getExtend(str,strLen,extend,mode,modeLen);int k=0,ans=0;while(k<strLen){if(extend[k]==modeLen){ans++;k=k+modeLen;}else{k=k+1;}}printf("%d\n",ans);}}

D:给你一串字符串，只能从前面加或后面加字符，问你最少加几次，可以让字符串有循环节,且循环次数大于1.

思路：kmp前缀函数的应用，若字符串有循环节，循环节长度一定是len-next[len]…

#include <cstdio>#include <cstring>#define size 100000char p[size+5];int  next[size+5];void prefix(char p[]){int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){if(k>0 && p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k;}}int main(){int T;scanf("%d",&T);while(T--){scanf("%s",p+1);prefix(p);int len=strlen(p+1);int count=len-next[len];printf("%d\n",(len%count==0 && len/count!=1)?0:count-len%count);}return 0;} 

E：字符串的所有前缀是否有循环节，若有输出循环次数

思路：前缀函数的应用。

ACcode：

#include <cstdio>#include <cstring>#define size 1000000char p[size+5];int next[size+5];void prefix(){int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){while(k>0 && p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k;}}void print(int n){for(int i=2;i<=n;i++){int t=i-next[i];if( i%t==0 && i/t!=1)printf("%d %d\n",i,i/t);}}int main(){int n;int k=1;while(scanf("%d",&n)==1&&n){scanf("%s",p+1);printf("Test case #%d\n",k++);prefix();print(n);printf("\n");}}

F:一个人用字符串A重复的写变成B:AAA…A，现在他把B随机分成2半.给你其中的一半，问最短可能的字符串a的长度

思路:同循环节问题

ACcode：

#include <cstdio>#include <cstring>#define size 1000000char p[size+5];int next[size+5];void prefix(char p[]){int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){while(k>0 && p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k; }}int main(){while(scanf("%s",p+1)!=EOF){prefix(p);int len=strlen(p+1);printf("%d\n",len-next[len]);}return 0;} 

G：字符串若存在循环节，输出循环次数

#include <cstdio>#include <cstring>#include <string>using namespace std;#define  T_SIZE 1000000+5 char t[T_SIZE];int next[T_SIZE];void prefix(char p[]){ int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){while(k>0 &&p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k;}}int main(){int len1;int len2;while(gets(t+1)){if(strcmp(t+1,".")==0)return 0;memset(next,0,sizeof(next));prefix(t);len1=strlen(t+1);len2=len1-next[len1];if(len1%len2==0) printf("%d\n",len1/len2);else printf("1\n");}return 0;}

H:给一字符串，找出所有既是前缀也是后缀的子串的位置

思路：ans=next[ans].递归找下去就是了。

#include <cstdio>#include <cstring>#define  T_SIZE 400000+5char p[T_SIZE];int next[T_SIZE],res[T_SIZE];void prefix(char p[]){int m=strlen(p+1);next[1]=0;for(int k=0,q=2;q<=m;q++){while(k>0&&p[k+1]!=p[q])k=next[k];if(p[k+1]==p[q])k++;next[q]=k;}}int main(){int i,j;while(gets(p+1)!=NULL){j=0;prefix(p);int len=strlen(p+1);int ans=next[len];res[j++]=len;while(ans>0){res[j++]=ans;ans=next[ans];}for(i=j-1;i>0;i--){printf("%d ",res[i]);}printf("%d\n",res[0]);}return 0;}