关于字符串的模板留底

都是从网上找的，交过题试过的应该没问题的模板。

KMP

//未优化的next函数
void getNext(char s[],int next[])
{
    int length=strlen(s);
    int i=0,j=-1;
    next[0]=-1;
    while(i<length)
    {
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else
            j=next[j];
    }
}

//已优化的next函数
void getNextval(char s[],int nextval[])
{
    int length=strlen(s);
    int i=0,j=-1;
    nextval[0]=-1;
    while(i<length)
    {
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            //next[i]=j;
            if (s[i]!=s[j])
                nextval[i]=j;
            else
                nextval[i]=nextval[j];
        }
        else
            j=nextval[j];
    }
}

//KMP
int KMP( char *t, char *s )   //s为主串，t为模式串
{
    int lenth = strlen(t);
    int len = strlen(s);
    getNextval( t, nextval );
    int i = 0, j = 0;
    while ( j < len )
    {
        if ( i == -1 || s[j] == t[i] )
        {
            ++i, ++j;
            if ( i == lenth ) return j;
        }
        else i = nextval[i];
    }
    return -1;
}

扩展KMP

#include<iostream>
#include<string>
using namespace std;
const int MM=100005;
int next[MM],extand[MM];
char S[MM],T[MM];
void GetNext(const char *T){
     int len=strlen(T),a=0;
     next[0]=len;
     while(a<len-1 && T[a]==T[a+1]) a++;
     next[1]=a;
     a=1;
     for(int k=2;k<len;k++){
         int p=a+next[a]-1,L=next[k-a];
         if( (k-1)+L >= p){
             int j = (p-k+1)>0 ? (p-k+1) : 0;
             while(k+j<len && T[k+j]==T[j]) j++;
             next[k]=j;
             a=k; 
         } 
         else
             next[k]=L; 
     } 
} 
void GetExtand(const char *S,const char *T){
     GetNext(T);
     int slen=strlen(S),tlen=strlen(T),a=0; 
     int MinLen = slen < tlen ? slen : tlen;
     while(a<MinLen && S[a]==T[a]) a++;
     extand[0]=a;
     a=0;
     for(int k=1;k<slen;k++){
         int p=a+extand[a]-1, L=next[k-a];
         if( (k-1)+L >= p){
             int j= (p-k+1) > 0 ? (p-k+1) : 0;
             while(k+j<slen && j<tlen && S[k+j]==T[j]) j++;
             extand[k]=j;
             a=k; 
         }
         else 
             extand[k]=L; 
     } 
} 
int main(){
    while(scanf("%s%s",S,T)==2){
         GetExtand(S,T);
         for(int i=0;i<strlen(T);i++)
             printf("%d ",next[i]);
         puts("");
         for(int i=0;i<strlen(S);i++)
             printf("%d ",extand[i]);
         puts(""); 
    } 
    return 0;
}

AC自动机

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>

using namespace std;

const int MAX_NODE = 1010;
const int CHILD_NUM = 4;
const int MAXN = 12;
const int INF = 1 << 30;

struct ACAutomaton
{
    int chd[MAX_NODE][CHILD_NUM]; //每个节点的儿子,即当前节点的状态转移
    int val[MAX_NODE];            //记录题目给的关键数据
    int fail[MAX_NODE];           //传说中的fail指针
    int Q[MAX_NODE<<1];           //队列,用于广度优先计算fail指针
    int ID[128];                  //字母对应的ID
    int sz;                       //已使用节点个数

    //初始化,计算字母对应的儿子ID,如:'a'->0 ... 'z'->25
    void Initialize()
    {
        fail[0] = 0;
        ID['A'] = 0;
        ID['G'] = 1;
        ID['C'] = 2;
        ID['T'] = 3;
        return;
    }
    //重新建树需先Reset
    void Reset()
    {
        memset(chd[0] , 0 , sizeof(chd[0]));
        val[0] = 0;
        sz = 1;
    }
    //将权值为key的字符串a插入到trie中
    void Insert(char *a,int key)
    {
        int p = 0;
        for ( ; *a ; a ++)
        {
            int c = ID[*a];
            if (!chd[p][c])
            {
                memset(chd[sz] , 0 , sizeof(chd[sz]));
                val[sz] = 0;
                chd[p][c] = sz ++;
            }
            p = chd[p][c];
        }
        val[p] = key;
    }
    //建立AC自动机,确定每个节点的权值以及状态转移
    void Construct()
    {
        int *s = Q , *e = Q;
        for (int i = 0 ; i < CHILD_NUM ; i ++)
        {
            if (chd[0][i])
            {
                fail[ chd[0][i] ] = 0;
                *e ++ = chd[0][i];
            }
        }
        while (s != e)
        {
            int u = *s++;
            for (int i = 0 ; i < CHILD_NUM ; i ++)
            {
                int &v = chd[u][i];
                if (v)
                {
                    *e ++ = v;
                    fail[v] = chd[ fail[u] ][i];
                    //以下一行代码要根据题目所给val的含义来写
                    val[v] |= val[ fail[v] ];
                }
                else
                {
                    v = chd[ fail[u] ][i];
                }
            }
        }
    }
} AC;

后缀数组

//rank从0开始
//sa从1开始,因为最后一个字符(最小的)排在第0位
//high从2开始,因为表示的是sa[i-1]和sa[i]
#define M 220000
int rank[M],sa[M],X[M],Y[M],high[M],init[M];
int buc[M];
void calhigh(int n) {
    int i , j , k = 0;
    for(i = 1 ; i <= n ; i++) rank[sa[i]] = i;
    for(i = 0 ; i < n ; high[rank[i++]] = k)
        for(k?k--:0 , j = sa[rank[i]-1] ; init[i+k] == init[j+k] ; k++);
}
bool cmp(int *r,int a,int b,int l) {
    return (r[a] == r[b] && r[a+l] == r[b+l]);
}
void suffix(int n,int m = 128) {
    int i , l , p , *x = X , *y = Y;
    for(i = 0 ; i < m ; i ++) buc[i] = 0;
    for(i = 0 ; i < n ; i ++) buc[ x[i] = init[i]  ] ++;
    for(i = 1 ; i < m ; i ++) buc[i] += buc[i-1];
    for(i = n - 1; i >= 0 ; i --) sa[ --buc[ x[i] ]] = i;
    for(l = 1,p = 1 ; p < n ; m = p , l *= 2) {
        p = 0;
        for(i = n-l ; i < n ; i ++) y[p++] = i;
        for(i = 0 ; i < n ; i ++) if(sa[i] >= l) y[p++] = sa[i] - l;
        for(i = 0 ; i < m ; i ++) buc[i] = 0;
        for(i = 0 ; i < n ; i ++) buc[ x[y[i]] ] ++;
        for(i = 1 ; i < m ; i ++) buc[i] += buc[i-1];
        for(i = n - 1; i >= 0 ; i --) sa[ --buc[ x[y[i]] ] ] = y[i];
        for(swap(x,y) , x[sa[0]] = 0 , i = 1 , p = 1 ; i < n ; i ++)
            x[ sa[i] ] = cmp(y,sa[i-1],sa[i],l) ? p-1 : p++;
    }
    calhigh(n-1);//后缀数组关键是求出high,所以求sa的时候顺便把rank和high求出来
}
 
 
//当需要反复询问两个后缀的最长公共前缀时用到RMQ
int Log[M];
int best[20][M];
void initRMQ(int n) {//初始化RMQ
    for(int i = 1; i <= n ; i ++) best[0][i] = high[i];
    for(int i = 1; i <= Log[n] ; i ++) {
        int limit = n - (1<<i) + 1;
        for(int j = 1; j <= limit ; j ++) {
            best[i][j] = min(best[i-1][j] , best[i-1][j+(1<<i>>1)]); 
        }
    }
}
int lcp(int a,int b) {//询问a,b后缀的最长公共前缀
    a = rank[a];    b = rank[b];
    if(a > b) swap(a,b);
    a ++;
    int t = Log[b - a + 1];
    return min(best[t][a] , best[t][b - (1<<t) + 1]);
}
 
 
int main() {
    //预处理每个数字的Log值,常数优化,用于RMQ
    Log[0] = -1;
    for(int i = 1; i <= M ; i ++) {
        Log[i] = (i&(i-1)) ? Log[i-1] : Log[i-1] + 1 ;
    }
    //*******************************************
    //    n为数组长度,下标0开始
    //    将初始数据,保存在init里,并且保证每个数字都比0大
    //    m = max{ init[i] } + 1
    //    一般情况下大多是字符操作,所以128足够了
    //*******************************************
    init[n] = 0;
    suffix(n+1,m);
 
    initRMQ(n);
}

 后缀自动机

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

#define N 2010
#define MAXQ 10010

//后缀自动机节点编号从1开始，ant为节点总数
//0号节点留作空余
struct Suffix_Automaton
{
    int F[N << 1],ant,last,ch[N << 1][26],step[N << 1];

    void init()
    {
        last = ant = 1;
        memset(F,0,sizeof(F));
        memset(ch,0,sizeof(ch));
        memset(step,0,sizeof(step));
    }

    void ins(int x)
    {
        int t = ++ant, pa = last;
        step[t] = step[last] + 1;
        last = t;
        for( ; pa && !ch[pa][x]; pa = F[pa] )
            ch[pa][x] = t;
        if( pa == 0 ) F[t] = 1;
        else if( step[pa] + 1 == step[ ch[pa][x] ] )
            F[t] = ch[pa][x];
        else
        {
            int nq = ++ant, q = ch[pa][x];
            memcpy( ch[nq], ch[q], sizeof(ch[nq]) );
            step[nq] = step[pa] + 1;
            F[nq] = F[q];
            F[q] = F[t] = nq;
            for( ; pa && ch[pa][x] == q; pa = F[pa] )
                ch[pa][x] = nq;
        }
    }
};