洛谷 5555 秩序魔咒

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又是一个广义 PAM 板题（

算每个状态在两个字符串分别的出现次数来判断是否公共，并用长度更新答案就完事了（

代码： 

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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 521634;
int n,m;
char s[N + 5];
int ans,cnt;
namespace PAM
{
    struct node
    {
        int ch[26];
        int fa,len,sz[2];
    } pam[N + 5];
    int las = 1,tot = 1;
    inline int init()
    {
        pam[1].len = -1,pam[0].fa = 1;
        return 0;
    }
    int Init = init();
    void insert(char *s,int i,int pos)
    {
        int cur = las,x = s[i] - 'a';
        for(;s[i - pam[cur].len - 1] ^ s[i];cur = pam[cur].fa);
        if(!pam[cur].ch[x])
        {
            int p = ++tot,q = pam[cur].fa;
            pam[p].len = pam[cur].len + 2;
            for(;s[i - pam[q].len - 1] ^ s[i];q = pam[q].fa);
            pam[p].fa = pam[q].ch[x],pam[cur].ch[x] = p;
        }
        ++pam[las = pam[cur].ch[x]].sz[pos];
    }
    inline void build()
    {
        for(register int i = tot;i > 1;--i)
        {
            pam[pam[i].fa].sz[0] += pam[i].sz[0],
            pam[pam[i].fa].sz[1] += pam[i].sz[1];
            if(pam[i].sz[0] && pam[i].sz[1])
                if(pam[i].len > ans)
                    ans = pam[i].len,cnt = 1;
                else if(pam[i].len == ans)
                    ++cnt;
        }
    }
}
int main()
{
    scanf("%d%d%s",&n,&m,s + 1);
    for(register int i = 1;i <= n;++i)
        PAM::insert(s,i,0);
    PAM::las = 1;
    scanf("%s",s + 1);
    for(register int i = 1;i <= m;++i)
        PAM::insert(s,i,1);
    PAM::build();
    printf("%d %d\n",ans,cnt);
}