ZOJ - 3661 pam

题意:给一个字符串,和每个字符代表的val,每个回文串的价值就是前半部分的val26进制%777777777,求价值第k小的回文串

题解:建个pam,然后dfs两边(0,1),统计价值sort一遍就好了

k爆int了,= =白wa了半天
//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 777777777

#define ld long double

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

//#define cd complex<double>

#define ull unsigned long long

#define base 1000000000000000000

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fin freopen("a.txt","r",stdin)

#define fout freopen("c.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

template<typename T>

inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>

inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;

char s[N];

ll res,f[N],val[30];

pll ans[N];

struct PAM{

    int ch[N][26],fail[N],len[N],s[N];

    ll cnt[N];

    int last,n,p;

    int newnode(int w)

    {

        for(int i=0;i<26;i++)ch[p][i] = 0;

        cnt[p] = 0;

        len[p] = w;

        return p++;

    }

    void init()

    {

        p = last = n = 0;

        newnode(0);

        newnode(-1);

        s[n] = -1;

        fail[0] = 1;

    }

    int getfail(int x)

    {

        while(s[n-len[x]-1] != s[n]) x = fail[x];

        return x;

    }

    void add(int c)

    {

        s[++n] = c;

        int cur = getfail(last);

        if(!ch[cur][c]){

            int now = newnode(len[cur]+2);

            fail[now] = ch[getfail(fail[cur])][c];

            ch[cur][c] = now;

        }

        last = ch[cur][c];

        cnt[last]++;

    }

    void cal()

    {

        for(int i=p-1;i>=0;i--)cnt[fail[i]]+=cnt[i];

    }

    void dfs(int u,int len,ll v)

    {

        for(int i=0;i<26;i++)

            if(ch[u][i])

            {

                ans[++res]=mp((v+f[len]*val[i]%mod)%mod,cnt[ch[u][i]]);

                dfs(ch[u][i],len+1,(v+f[len]*val[i]%mod)%mod);

            }

    }

}pam;

int main()

{

//    fin;fout;

    f[0]=1;

    for(int i=1;i<N;i++)f[i]=f[i-1]*26ll%mod;

    int T;scanf("%d",&T);

    while(T--)

    {

        pam.init();

        int n,m;scanf("%d%d%s",&n,&m,s+1);

        for(int i=1;i<=n;i++)pam.add(s[i]-'a');

        pam.cal();

        while(m--)

        {

            ll k;scanf("%lld",&k);

            for(int i=0;i<26;i++)scanf("%lld",&val[i]);

            res=0;

            pam.dfs(0,0,0);pam.dfs(1,0,0);

//            printf("%d\n",res);

            sort(ans+1,ans+1+res);

            ll co=0;

            for(int i=1;i<=res;i++)

            {

                co+=ans[i].se;

                if(co>=k)

                {

                    printf("%lld\n",ans[i].fi);

                    break;

                }

            }

        }

        puts("");

    }

    return 0;

}

/********************

10

78 3

abaabbabababcddajkjdlspeoiabaabbabababcddajkjdlspeoiabaabbabababcddajkjdlspeoi

23 25 25 22 23 24 25 16 17 18 19 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

********************/
巴特西

ZOJ - 3661 pam

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