[SPOJ] (1812) Longest Common Substring ---- SAM(最长公共子串)

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题目传送门

做法:

  • 后缀家族的基本应用,之前学习了后缀数组的解法,后缀自动机的思想如下:
  • 给第一个串建立自动机,第二个串在自动机上匹配,通过tran数组,如果匹配就len+1,
    否则通过slink数组(fa)向前跳,直到能继续匹配。

AC代码:

#include<bits/stdc++.h>
using namespace std;

#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define abs(x)      ((x)<0 ? -(x) : x)
#define all(x)      x.begin(),x.end()
#define mk(x,y)     make_pair(x,y)
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)

typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 3e5+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;

char s[maxn],b[maxn];
struct SuffixAutoMation{
    int root,cnt,last,len;
    int trans[maxn<<1][26],slink[maxn<<1],l[maxn<<1];
    inline void init()
    {
        root = cnt = last = 1;
        memset(trans,0,sizeof(trans));
        slink[1] = l[1] = 0;
    }
    inline void add(int x)
    {
        int p = last,np = ++cnt;last = np;l[np] = l[p]+1;
        for(;p && !trans[p][x];p = slink[p]) trans[p][x] = np;
        if(!p) slink[np] = 1;
        else{
            int q = trans[p][x];
            if(l[p]+1 == l[q]) slink[np] = q;
            else{
                int nq = ++cnt;l[nq] = l[p]+1;
                memcpy(trans[nq],trans[q],sizeof(trans[q]));
                slink[nq] = slink[q];
                slink[q] = slink[np] = nq;
                for(;trans[p][x] == q; q = slink[q]) trans[p][x] = nq;
            }
        }
    }
    inline void build()
    {
        init();
        scanf("%s",s+1);
        scanf("%s",b+1);
        len = strlen(s+1);
        for(int i=1;i<=len;i++) add(s[i]-'a');
    }
    inline int Find() //第二个串与第一个串匹配
    {
        int m = strlen(b+1);
        int res = 0,ans = 0,p = root;
        for(int i=1;i<=m;i++)
        {
            int x = b[i]-'a';
            if(trans[p][x]) res++,p = trans[p][x];
            else{
                for(;p && !trans[p][x];p = slink[p]);
                if(!p) res = 0,p = root;
                else res = l[p]+1, p = trans[p][x];
            }
            ans = max(ans,res);
        }
        return ans;
    }
}sam;
int main()
{
    // fin;
    // IO;
    sam.build();
    cout<<sam.Find()<<endl;
    return 0;
}

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转载自blog.csdn.net/m0_37624640/article/details/83307829