题目链接

思路

\(SAM\)练手题，将原串重复一遍插入到\(SAM\)中，然后贪心走长度为n的一个路径即可。
不用担心会直接走到终点，根据\(SAM\)的构造方式可以发现会先走到前面的路径。

代码

/*
* @Author: wxyww
* @Date:   2019-07-11 11:09:25
* @Last Modified time: 2019-07-11 11:20:42
*/
#include<cstdio>
#include<map>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 600000 + 100;
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*10+c-'0';
        c=getchar();
    }
    return x*f;
}
struct node {
    int len,fa;
    map<int,int>ch;
}SAM[N << 1];
int lst = 1,tot = 1;
void add(int c) {
    int cur = ++tot,p = lst;
    SAM[cur].len = SAM[p].len + 1;
    while(p && !SAM[p].ch[c]) {
        SAM[p].ch[c] = cur;
        p = SAM[p].fa;
    }
    if(!p) SAM[cur].fa = 1;
    else {
        int q = SAM[p].ch[c];
        if(SAM[q].len == SAM[p].len + 1) SAM[cur].fa = q;
        else {
            int clone = ++tot;
            SAM[clone] = SAM[q];
            SAM[clone].len = SAM[p].len + 1;
            while(p && SAM[p].ch[c] == q) {
                SAM[p].ch[c] = clone;
                p = SAM[p].fa;
            }
            SAM[q].fa = SAM[cur].fa = clone;
        }
    }
    lst = cur;
}
int a[N];
int main() {
    int n = read();
    for(int i = 1;i <= n;++i) a[i] = read();
    for(int i = 1;i <= n;++i) add(a[i]);
    for(int i = 1;i <= n;++i) add(a[i]);
    int p = 1;
    for(int i = 1;i <= n;++i) {
        printf("%d ",SAM[p].ch.begin() -> first);
        p = SAM[p].ch.begin() -> second;
    }
    return 0;
}

luogu1368 工艺

原文：https://www.cnblogs.com/wxyww/p/luogu1368.html