spoj LCS2 【後綴自動機】

好捨不得大家……心疼……

題目大意：把上一題的兩個串改成多個串

對每一個串進行一次操作，記得把每個點的值傳給自己的par
最後取個min就好了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define N 200005
#define INF 1000000000
using namespace std;

int n,m,last = 1,tot = 1,p,q,np,nq,ans;
int son[N][26],par[N];
int P[N],num[N],mx[N],mv[N];
char s[N];

int new_node(int x)
{
    mx[++ tot] = x;
    return tot;
}

void add(int x)
{
    p = last;
    np = new_node(mx[p] + 1);
    for (;p && !son[p][x];p = par[p]) son[p][x] = np;
    if (!p) par[np] = 1;
    else
    {
        q = son[p][x];
        if (mx[q] == mx[p] + 1) par[np] = q;
        else
        {
            nq = new_node(mx[p] + 1);
            memcpy(son[nq],son[q],sizeof(son[nq]));
            par[nq] = par[q],par[q] = par[np] = nq;
            for (;son[p][x] == q;p = par[p]) son[p][x] = nq;
        }
    }
    last = np;
}

int main()
{
    scanf("%s",s + 1),n = strlen(s + 1);
    for (int i = 1;i <= n;i ++) add(s[i] - 'a');
    for (int i = 1;i <= tot;i ++) mv[mx[i]] ++;
    for (int i = 1;i <= n;i ++) mv[i] += mv[i - 1];
    for (int i = 1;i <= tot;i ++) P[mv[mx[i]] --] = i;
    memset(mv,63,sizeof(mv));
    while (~scanf("%s",s + 1))
    {
        int len = strlen(s + 1);
        for (int i = 1,p = 1,l = 0;i <= len;i ++)
        {
            int x = s[i] - 'a';
            if (son[p][x]) l ++,p = son[p][x];
            else
            {
                for (;p && !son[p][x];p = par[p]);
                if (!p) l = 0,p = 1;
                else l = mx[p] + 1,p = son[p][x];
            }
            num[p] = max(num[p],l);
        }
        for (int i = tot;i > 1;i --)
        {
            int x = P[i];
            mv[x] = min(mv[x],num[x]);
            if (par[x] && num[x]) num[par[x]] = mx[par[x]];
            num[x] = 0;
        }
    }
    for (int i = 2;i <= tot;i ++) ans = max(ans,mv[i]);
    cout << ans << endl;

    return 0;
}