P1368 工艺

题目

P1368 工艺
你好水题

做法

备份一下原数组，然后插入，直接输出贪心每位最小的就好了

My complete code

再见水题

#include<map>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef int LL;
const LL maxn=1000000;
inline LL Read(){
    LL x(0),f(1);char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
LL n,nod=1,lst=1;
LL a[maxn];
struct SAM{
    LL fail,len;
    map<LL,LL> son;
}S[maxn];
inline void Insert(LL c){
    LL np=++nod,p=lst; lst=nod;
    S[np].len=S[p].len+1;
    while(p&&S[p].son[c]==0){
        S[p].son[c]=np;
        p=S[p].fail;
    }
    if(!p)
        S[np].fail=1;
    else{
        LL q=S[p].son[c];
        if(S[q].len==S[p].len+1)
            S[np].fail=q;
        else{
            LL nq=++nod; S[nq].len=S[p].len+1;
            S[nq].son=S[q].son;
            S[nq].fail=S[q].fail;
            S[q].fail=S[np].fail=nq;
            while(p&&S[p].son[c]==q){
                S[p].son[c]=nq;
                p=S[p].fail;
            }
        }
    }
}
int main(){
    n=Read();
    for(LL i=1;i<=n;++i)
        a[i]=Read();
    for(LL i=1;i<n;++i)
        a[i+n]=a[i];
    for(LL i=1;i<2*n;++i)
        Insert(a[i]);
    LL now(1);
    for(LL i=1;i<=n;++i){
        printf("%d ",S[now].son.begin()->first);
        now=S[now].son.begin()->second;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/y2823774827y/p/10317546.html