Practice II 字符串

本来想做数论的……但是别的dalao都在做制胡窜

所以……

Chapter I KMP

KMP 最关键的不是这个半暴力的单模匹配

而是这个nxt数组 经常出一些奇怪的题 尤其是循环节可以直接由T-nxt[T]得到……神啊

总之记住nxt就是最长公共前后缀中前缀的尾指针

T1 poj3461 Oulipo

传送门

Time cost: 10min

纯纯的板子 真没啥可说的

就是每次清空nxt

Code:

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1000005;
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define ms(a,b) memset(a,b,sizeof a)

char s[N],t[N];
int S,T;
int nxt[N];
int ans;
void clr(){ms(nxt,0),ans = 0;}
void gnxt() {
    int tmp = 0;
    rep(i,2,T) {
    while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp];
    if(t[tmp+1] == t[i]) nxt[i] = ++tmp;
    }
}
void KMP() {
    int tmp = 0;
    rep(i,1,S) {
    while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp];
    if(t[tmp+1] == s[i]) {
        tmp++;
        if(tmp == T) ans++;
    }
    }
}


int main() {
    int q;
    scanf("%d",&q);
    while(q--) {
    clr();
    scanf("%s",t+1),scanf("%s",s+1);
    S = strlen(s+1),T = strlen(t+1);
    gnxt();
    KMP();
    printf("%d
",ans);
    }
}

T2 poj 2406 Power strings

传送门

Time cost:55min

之前曾经遇到过……最后暴力滚粗

今天算是靠着打表理解了

由于nxt定义的时候刨去了串本身是自己的最长公共前后缀

在最后一次求nxt的时候 我们考虑被nxt[T]抛弃的部分（记为len）

如果这一段能整除T，那么考虑第二段len长的串

它与第一段len长的串是相等的

同理可以推到T/len倍

同时 nxt取最大值 故T-nxt[T]取最小

所以如果T是T-nxt[T]的整数倍 那么T-nxt[T]就是最小循环节

Code:

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1000005;
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define ms(a,b) memset(a,b,sizeof a)

char s[N],t[N];
int S,T;
int nxt[N];
int ans;
void clr(){ms(nxt,0),ans = 0;}
void gnxt() {
    int tmp = 0;
    rep(i,2,T) {
    while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp];
    if(t[tmp+1] == t[i]) nxt[i] = ++tmp;
    }
}
void KMP() {
    int tmp = 0;
    rep(i,1,S) {
    while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp];
    if(t[tmp+1] == s[i]) {
        tmp++;
        if(tmp == T) ans++;
    }
    }
}


int main() {
    while(1) {
    clr();
    scanf("%s",t+1);
    T = strlen(t+1);
    if(T == 1 && t[1] == '.') return 0;
    gnxt();
    if(T % (T - nxt[T]) == 0) printf("%d
",T/(T-nxt[T]));
    else puts("1");
    }
}

T3 CF526D Om Nom and Necklace

传送门

Time cost:45min

有了前一道题做铺垫 还是比较容易想到的

交叉没法做 我们视AB为循环节 

由于k给定 所以通过i和k可以求出一个循环节长度范围

如果这个长度是由nxt求出的最短循环节的整数倍就OK

实现的时候是 除len然后判断合法区间是否存在 这样就可以自动减掉后面剩的一段A

但是要注意 A或B为空的情况就是分成k段或者k+1段纯循环节（没有后缀） 可以特判

（实际A为空已经处理过了）

Code:

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1000005;
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define ms(a,b) memset(a,b,sizeof a)

char s[N],t[N];
int S,T;
int nxt[N];
int ans;
void clr(){ms(nxt,0),ans = 0;}
void gnxt() {
    int tmp = 0;
    rep(i,2,T) {
    while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp];
    if(t[tmp+1] == t[i]) nxt[i] = ++tmp;
    }
}
void KMP() {
    int tmp = 0;
    rep(i,1,S) {
    while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp];
    if(t[tmp+1] == s[i]) {
        tmp++;
        if(tmp == T) ans++;
    }
    }
}
int n,k;

int main() {
    scanf("%d%d",&n,&k);
    scanf("%s",t+1);
    T = n;
    gnxt();
    rep(i,1,T) {
    int len = i - nxt[i];
    if(i % (k+1) == 0 && (i / (k+1)) % len == 0) putchar('1');//lenb==0
    else {
        //circular subsequence range
        int upp = i / k,down = i / (k + 1) + 1;
        upp = upp / len;
        down = (down + len - 1) / len;
        if(upp >= down) putchar('1');
        else putchar('0');
    }
    }
    puts("");
    return 0;
}

Chapter II Trie

字典树 比较重要的地方就是空间

一定要考虑好空间问题不要超限

其他的话还是比较好理解的 当做一棵26叉树

T1 poj3630 Phone list

Time cost: 25min

经典的Trie树板子题

先把所有串压进去然后再每一个串查

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
inline int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 10005;
const int M = 12;
#define ms(a) memset(a,0,sizeof(a));
//head
char s[N][M];
struct trie {
    int nxt[N * M][12],val[N * M];
    int cnt;
    inline void clear() {
        ms(nxt);
        ms(val);
        cnt = 0;
    }
    inline void build(char p[]) {
        int now = 0;
        rep(i,0,strlen(p) - 1) {
            if(!nxt[now][p[i] - '0'])
                nxt[now][p[i] - '0'] = ++cnt;
            now = nxt[now][p[i] - '0'];
        }
        val[now] = 1;
    }
    inline bool query(char p[]) {
        int now = 0;
        rep(i,0,strlen(p) - 1) {
            if(val[now]) return 0;
            now = nxt[now][p[i] - '0'];
        }
        return 1;
    }
} T;

int main() {
    int q = read();
    while(q--) {
        bool flag = 1;
        ms(s);
        T.clear();
        int n = read();
        rep(i,1,n) {
            scanf("%s",s[i]);
            T.build(s[i]);
        }
        rep(i,1,n) {
            flag = T.query(s[i]);
            if(flag == 0) break;
        }
        flag ? printf("YES
") : printf("NO
") ;
    }
    return 0;
}

T2 poj2945 Find the clones

Time cost:25min

题目要求输出被复制x次的有多少种

依然先把所有串建好树

那么就在query的时候开一个res表示val[now]的有多少个

注意每次找到的时候要把val清空

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
inline int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 20005;
const int M = 25;
#define ms(a) memset(a,0,sizeof(a));
//head
char s[N][M];
struct trie {
    int nxt[N * M][5],val[N * M];
    int res[N];
    int cnt;
    inline void clear() {
        ms(nxt);
        ms(val);
        ms(res);
        cnt = 0;
    }
    inline int cng(char c) {
        if(c == 'A') return 0;
        if(c == 'G') return 1;
        if(c == 'T') return 2;
        if(c == 'C') return 3;
        else return 4;
    }
    inline void build(char p[]) {
        int now = 0;
        rep(i,0,strlen(p) - 1) {
            int c = cng(p[i]);
            if(!nxt[now][c])
                nxt[now][c] = ++cnt;
            now = nxt[now][c];
        }
        val[now]++;
    }
    inline void query(char p[]) {
        int now = 0;
        rep(i,0,strlen(p) - 1) {
            int c = cng(p[i]);
            now = nxt[now][c];
        }
        res[val[now]]++;
        val[now] = 0;
    }
} T;

int main() {
    while(1) {
        int n = read();
        int m = read();
        if(!n && !m) break;
        T.clear();
        ms(s);
        rep(i,1,n) {
            scanf("%s",s[i]);
            T.build(s[i]);
        }
        rep(i,1,n) T.query(s[i]);
        rep(i,1,n) {
            printf("%d
",T.res[i]);
        }
    }
    return 0;
}

T3 poj1816 Wild words

Time cost:65min

好像这题有个名字叫模糊查询……

?的话就额外开一个节点 查的时候单独查

*的话就暴搜除掉的位数……

注意S* 也能匹配S

所以搜*的时候要在搜结束节点之前

然后就是空间……这道题nxt数组开1e5*30*30都不行 每个节点存答案的时候必须用vector

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
const int N = 100001;
const int M = 28;
//HEAD
int m,n;
vector<int> mem[N*M];
int ans[N],top;
int nxt[N][M],cnt;
void clr() {
    cnt = 0;
    ms(nxt,0);
}
int cng(char op) {
    if(op == '*') return 26;
    if(op == '?') return 27;
    return op - 'a';
}

char p[35];
int len;
void ins(int idx) {
    int cur = 0;
    rep(i,0,len - 1) {
    if(!nxt[cur][cng(p[i])])
        nxt[cur][cng(p[i])] = ++cnt;
    cur = nxt[cur][cng(p[i])];
    }
    mem[cur].push_back(idx);
}

void query(int cur,int pos) {
    if(nxt[cur][26]) 
    rep(i,pos,len) query(nxt[cur][26],i);
    
    if(pos == len) {
    int sze = mem[cur].size();
    rep(i,0,sze - 1)
        ans[++top] = mem[cur][i];
    return;
    }
    if(nxt[cur][cng(p[pos])])
    query(nxt[cur][cng(p[pos])],pos+1);
    if(nxt[cur][27])
    query(nxt[cur][27],pos+1);
}

int main() {
    scanf("%d%d",&n,&m);
    clr();
    rep(i,1,n) {
    scanf("%s",p);
    len = strlen(p);
    ins(i-1);
    }
    while(m--) {
    scanf("%s",p);
    top = 0;
    len = strlen(p);
    query(0,0);
    if(!top) {
        puts("Not match");
        continue;
    }
    sort(ans+1,ans+1+top);
    top = unique(ans+1,ans+1+top) - ans - 1;
    rep(i,1,top) printf("%d ",ans[i]);
    puts("");
    }
    return 0;
}

T4 CF633C Spy Syndrome 2

传送门

Time cost:105min

发现做字符串时间格外长……还是不熟练

一开始想着先反过来再匹配……

事实证明考虑前一位是否匹配的时候如果是正的还是要遍历一遍

所以直接就倒着做就行

O(n^2)可以做出类似链式前向星的东西 然后从最后一位输出就可以了

注意标记初值要设成-1 

然后递归输出的时候注意顺序不要反了就行

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 100005;
//head
int n,m;
char S[10005];
char p[N][1005];
int cng(char op){return op >= 'a' ? op - 'a' : op - 'A';}

int nxt[1000005][26],val[1000005],tot;
void ins(char *s,int id) {
    int cur = 0;
    int len = strlen(s);
    rep(i,0,len-1) {
        int c = cng(s[i]);
        if(!nxt[cur][c]) nxt[cur][c] = ++tot;
        cur = nxt[cur][c];
    }
    val[cur] = id;
}

int pre[N],ans[N];
void query(int x) {
    int cur = 0;
    per(i,x,0) {
        int c = cng(S[i]);
        cur = nxt[cur][c];
        if(!cur) return;
        if(val[cur]) {
            if(ans[i-1] || !i) {
                pre[x] = i-1;
                ans[x] = val[cur];
                return;
            }
        }
    }
}

void output(int x) {
    if(~pre[x]) output(pre[x]);
    printf("%s ",p[ans[x]]);
}

int main() {
    n = read();
    scanf("%s",S);
    m = read();
    rep(i,1,m) {
        scanf("%s",p[i]);
        ins(p[i],i);
    }
    ms(pre,-1);
    rep(i,0,n-1) query(i);
    output(n-1),puts("");
    return 0;
}

Chapter III AC自动机

我太菜了……板子调了好久……

总体来讲 AC自动机与KMP相同的是 fail很重要

不同的是 这个匹配也很重要

都可以拿来出题

总体来讲 沿着fail的路线也可以搞事情（poj4052），沿着trie树向上走也可以做标记

总之 只有节点维护标记 节点之间的联系只有trie边和fail 考虑清楚就行了

T1 poj 4052 Hrinity

Time cost:95min

这么水一个题我做这么长时间……太菜了

题意大概是给出一堆模式串（还有缩写）在文本串里找

其中被包含的子串不计数

不能在文本串里去掉因为可能剩下的还能匹配小串

所以就第一次先把所有匹配了的串都打上标记 然后再扫一遍把被普通节点覆盖的终止节点去掉

注意：

1.不要把自己去了

2.转换的时候用strcpy比较方便 但别用反了……

3.去掉小串之前要先把访问数组清空

Code:

Source Code

Problem: 4052        User: dvb
Memory: 36408K        Time: 594MS
Language: G++        Result: Accepted
Source Code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define itn int
#define inf 2147483647
#define eps 1e-9
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
int n;
char p[2505][1105],S[5100005];
char P[5100005];

void cng(char s[]) {
    int len = strlen(s);
    int cur = 0;
    rep(i,0,len-1) {
    if(s[i] == '[') {i++;
        int num = 0;
        char op;
        while(s[i] != ']') {
        if(s[i] >= '0' && s[i] <= '9') num = num * 10 + s[i] - '0';
        else op = s[i];
        i++;
        }
        rep(i,1,num) P[cur++] = op;
    }
    else P[cur++] = s[i];
    }
    P[cur] = '';
    strcpy(s,P);
}


struct node {
    int val,fail;
    int nxt[26];
    void init() {
    ms(nxt,0),val = fail = 0;
    }
}t[2750005];
int tot = 1;

void ins(char *s,int id) {
    int len = strlen(s);
    int cur = 1;
    rep(i,0,len-1) {
    int c = s[i] - 'A';
    if(!t[cur].nxt[c])
        t[cur].nxt[c] = ++tot,t[tot].init();
    cur = t[cur].nxt[c];
    }
    t[cur].val = id;
}


void getfail() {
    rep(i,0,25) t[0].nxt[i] = 1;
    queue<int> q;
    q.push(1);
    while(!q.empty()) {
    int u = q.front(),v,w;
    q.pop();
    rep(i,0,25) {
        v = t[u].fail;
        while(!t[v].nxt[i]) v = t[v].fail;
        v = t[v].nxt[i];
        w = t[u].nxt[i];
        if(w) t[w].fail = v,q.push(w);
        else t[u].nxt[i] = v;
    }
    }
}


bool vis[2750005];
bool flag[2505];
void query() {
    int len = strlen(S);
    int cur = 1;
    rep(i,0,len-1) {
    cur = t[cur].nxt[S[i]-'A'];
    if(t[cur].val) {
        flag[t[cur].val] = 1;
        continue;
    }
    int tmp = cur;
    while(tmp && !vis[tmp]) {
        vis[tmp] = 1;
        flag[t[tmp].val] = 1;
        tmp = t[tmp].fail;
    }
    }
}

void del(char *s,int id) {
    int len = strlen(s);
    int cur = 1;
    rep(i,0,len-1) {
    cur = t[cur].nxt[s[i]-'A'];
    int tmp = cur;
    while(tmp && !vis[tmp]) {
        if(t[tmp].val && t[tmp].val ^ id) {
        vis[tmp] = 1;
        flag[t[tmp].val] = 0;
        }
        tmp = t[tmp].fail;
    }
    }
}

void clr() {
    rep(i,1,tot) t[i].init();
    tot = 1;
    ms(flag,0);
}

int main() {
    int T = read();
    while(T--) {
    clr();
    
    n = read();
    rep(i,1,n) {
        scanf("%s",p[i]);
        cng(p[i]);
        ins(p[i],i);
    }
    
    getfail();
    
    scanf("%s",S);
    cng(S);
    ms(vis,0),query();
    
    ms(vis,0);
    rep(i,1,n) 
        if(flag[i]) del(p[i],i);
    
    int ans = 0;
    rep(i,1,n) ans += flag[i];
    printf("%d
",ans);
    
    }
    return 0;
}

Chapter IV manacher

回文串...看起来比较不常见 但是出了就得会 

听说后面有个回文自动机? 看到了再说吧...

T1  poj3974 Palindrome 

Time cost: 35min

基础的回文串题 调了一下之前的板子 换了种写法

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 1000005;
//head
char s[N<<1],t[N];
int R[N<<1];
int n,m,T,maxx;
void manacher() {
    int id = 0,pos = 0,r = 0;
    rep(i,1,n) {
    if(pos > i) r = min(R[id * 2 - i],pos - i);
    else r = 1;
    while(s[i - r] == s[i + r]) r++;
    if(r + i > pos) {
        pos = r + i;
        id = i;
    }
    R[i] = r;
    }
}

void clr() {
    ms(R,0);
    maxx = 0;
}
int main() {
    while(scanf("%s",t+1),t[1] != 'E') {
    clr();
    m = strlen(t+1);
    s[n = 0] = 0;
    rep(i,1,m) {
        s[++n] = '#';
        s[++n] = t[i];
    }
    s[++n] = '#';
    s[n+1] = 1;
    manacher();
    rep(i,1,n) maxx = max(maxx,R[i]);
    printf("Case %d: %d
",++T,maxx - 1);
    }
    return 0;
}  

T2 luogu P1659 [国家集训队]拉拉队排练 

传送门

Time cost:45min

发现自己还是不太明白原理....

这道题做法就是找出所有长为奇数的回文串

发现长的回文串一定包含较短的

所以求一遍最长长度的后缀和就是真正的个数

数据范围较大 使用快速幂

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 1000005;
const ll mod = 19930726ll;
//head
char s[N];
int r[N];
ll n,k,ans = 1;
ll num[N];
void manacher() {
    int id = 0,pos = 0,x = 0;
    rep(i,1,n) {
    if(pos > i) x = min(r[(id<<1)-i] , pos - i);
    else x = 1;
    while(s[i+x] == s[i-x]) x++;
    if(x + i > pos) pos = x + i,id = i;
    r[i] = x;
    num[r[i]-1<<1|1]++;
    }
}
ll ksm(ll a,ll b) {
    ll tmp = 1;
    while(b) {
    if(b & 1) tmp = tmp * a % mod;
    a = a * a % mod;
    b >>= 1;
    }
    return tmp;
}


int main() {
    n = read(),k = read();
    scanf("%s",s+1);
    s[0] = '&';
    manacher();
    per(i,n+1>>1,1) {
    ll l = i-1<<1|1;
    num[l] += num[l+2];
    if(k >= num[l]) {
        ans = (ans * ksm(l,(ll)num[l])) % mod;
        k -= num[l];
    } else {
        ans = (ans * ksm(l,k)) % mod;
        printf("%lld
",ans);
        k = 0;
        break;
    }
    }
    if(k) puts("-1");
    return 0;
}  

T3 luogu  P4555 [国家集训队]最长双回文串

传送门

Time cost:85min

难度明显上来了

不能直接暴力 考虑枚举中间的连接点

然后思路就是找以当前点为左右端点的最远另一端点（记为L，R）

注意真正的回文串长度是所求回文半径长度-1

所以就在manacher里面找到最长回文串的时候求出其左右端点的答案

注意这个答案还可以被L（R）更新

所以我们找到所有的下标为‘#’（插入的特殊字符）的位置互相更新（具体见代码main）

复杂度O(n)

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 200005;
//head
int n,maxx;
char s[N],t[N];
int r[N],L[N],R[N];
void manacher() {
    int pos = 0,maxr = 0,x = 0;
    rep(i,1,n) {
    if(maxr > i) x = min(r[(pos<<1)-i],maxr-i);
    else x = 1;
    while(s[i + x] == s[i - x]) x++;
    if(i + x > maxr) maxr = i+x,pos = i;
    r[i] = x;
    L[i+r[i]-1] = max(L[i+r[i]-1],r[i]-1);
    R[i-r[i]+1] = max(R[i-r[i]+1],r[i]-1);
    }
}

int main() {
    scanf("%s",t+1);
    s[++n] = 0;
    int l = strlen(t+1);
    rep(i,1,l) {
    s[++n] = '#';
    s[++n] = t[i];
    }
    s[++n] = '#';
    s[++n] = '%';
    manacher();
    rep(i,1,n>>1) R[i<<1] = max(R[i<<1],R[i-1<<1]-2);
    per(i,n>>1,1) L[i<<1] = max(L[i<<1],L[i+1<<1]-2);
    rep(i,1,n>>1) if(R[i<<1] && L[i<<1]) maxx = max(maxx,L[i<<1] + R[i<<1]);
    printf("%d
",maxx);
    return 0;
}

To be continued...

https://www.luogu.org/problemnew/show/P1659