UVA 719 Glass Beads SAM

最新推荐文章于 2021-09-28 22:14:30 发布

原创最新推荐文章于 2021-09-28 22:14:30 发布 · 158 阅读

0 ·

CC 4.0 BY-SA版权

ACM 同时被 2 个专栏收录

455 篇文章

订阅专栏

后缀自动机SAM

2 篇文章

订阅专栏

本文介绍了一种使用SAM（Suffix Array Machine）算法解决特定字符串问题的方法。通过构建2倍字符串并存储到SAM中，采用字典序贪心策略找到最小表示法的起始位置。文章提供了完整的C++代码实现。

题意给你多组n 每组n个字符串输出最小表示法开始的下标
做法我们把2*字符串存入SAM中然后只要按字典序贪心的取最小的返回maxlen[now]-len+1即使需要的答案

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;


const int MAX_N = 100010;
char str[MAX_N];
int len;
struct SAM
{
    int fa[MAX_N],ch[MAX_N][26],maxlen[MAX_N],tot,last;
    void init()
    {
        tot = last = 1;maxlen[1] = fa[1] = 0;
        memset(ch[1],0,sizeof(ch[1]));
    }
    void add(int x)
    {
        int np = ++tot,p = last;last = np;
        maxlen[np] = maxlen[p] + 1;
        memset(ch[np],0,sizeof(ch[np]));
        while(p&&!ch[p][x]) ch[p][x] = np,p = fa[p];
        if(!p) fa[np] = 1;
        else
        {
            int q = ch[p][x];
            if(maxlen[q]==maxlen[p]+1) fa[np] = q;
            else
            {
                int nq = ++tot;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                fa[nq] = fa[q],fa[np] = fa[q] = nq;
                maxlen[nq] = maxlen[p] + 1;
                while(p&&ch[p][x]==q) ch[p][x]=nq,p = fa[p];
            }
        }
    }
    void solve()
    {
        int NOW = 1;
        for(int i = 0;i<len;++i)
        {
            for(int j = 0;j<26;j++)
            {
                if(ch[NOW][j])
                {
                    NOW = ch[NOW][j];
                    break;
                }
            }
        }
        printf("%d\n",maxlen[NOW]-len+1);
    }
}S;

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    while(~scanf("%d",&n))
    {
        while(n--)
        {
            S.init();
            scanf("%s",str);
            len = strlen(str);
            for(int i = 0; i<len; ++i)
                S.add(str[i]-'a');
            for(int i = 0; i<len; ++i)
                S.add(str[i]-'a');
            S.solve();
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}