BZOJ 2160: 拉拉队排练

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双倍经验题

manacher/palindromic tree

对于manacher的话显然回文串的长度最多只有n种，对于每个奇数长度的位置显然长度为1,3,5.....r[i]的串的个数都+1，区间加减什么的差分一下前缀和搞一搞就好了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<cmath>using namespace std;#define rep(i,l,r) for(int i=l;i<=r;i++)#define per(i,r,l) for(int i=r;i>=l;i--)#define mmt(a,v) memset(a,v,sizeof(a))const int N=1000000+5;const int p=19930726;typedef long long ll;char s[N<<1],t[N];int r[N<<1],n;ll cnt[N],k;void manacher(){    int m=n<<1|1;    rep(i,1,m)    if(i&1)s[i]='#';    else s[i]=t[i>>1];    int id,mx=0;    rep(i,1,m){        if(mx>i)r[i]=min(mx-i,r[2*id-i]);        else r[i]=1;        while(i-r[i]>=1&&i+r[i]<=m&&s[i-r[i]]==s[i+r[i]])r[i]++;        if(~r[i]&1)cnt[(r[i]>>1)]++;        if(i+r[i]>mx)mx=i+r[i],id=i;    }}ll qmul(ll a,ll b){    ll ans=1;    for(;b;b>>=1,a=a*a%p)if(b&1)ans=ans*a%p;    return ans;}int solve(){    manacher();    ll ans=1;    per(i,n,1){        cnt[i]+=cnt[i+1];        if(cnt[i]<=k){            k-=cnt[i];            (ans*=qmul(2*i-1,cnt[i]))%=p;        }else (ans*=qmul(2*i-1,k))%=p,k=0;        if(!k)return ans;    }    return -1;}int main(){    //freopen("a.in","r",stdin);    scanf("%d%lld",&n,&k);    scanf("%s",t+1);    printf("%d\n",solve());    return 0;}

回文树就是裸题了，lazytag打一打，逆序更新一下就好了（由于我比较懒就没基数排序QAQ结果就是nlogn了，好像慢了5倍左右）

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<cmath>using namespace std;#define rep(i,l,r) for(int i=l;i<=r;i++)#define per(i,r,l) for(int i=r;i>=l;i--)#define mmt(a,v) memset(a,v,sizeof(a))const int N=1000000+5;const int p=19930726;typedef long long ll;ll qmul(ll a,ll b){    ll ans=1;    for(;b;b>>=1,a=a*a%p)if(b&1)ans=ans*a%p;    return ans;}struct Node{    int len,suf,ch[26];    ll sum;    bool operator < (const Node &x)const{        return len>x.len;    }}tr[N];int last,node;void init(){last=node=2;tr[tr[tr[2].suf=1].suf=1].len=-1;}char s[N];void add(int i){    int cur=last,c=s[i]-'a';    while(s[i-tr[cur].len-1]!=s[i])cur=tr[cur].suf;    if(!tr[cur].ch[c]){        tr[last=++node].len=tr[cur].len+2;tr[cur].ch[c]=last;        int tmp=tr[cur].suf;        while(s[i-tr[tmp].len-1]!=s[i])tmp=tr[tmp].suf;        tr[last].suf=tr[last].len==1?2:tr[tmp].ch[c];    }    last=tr[cur].ch[c];tr[last].sum++;}int n;ll k;int solve(){    ll ans=1;    rep(i,1,node){        if(tr[i].len<=0)continue;        if(~tr[i].len&1)continue;        if(tr[i].sum<=k)        (ans*=qmul(tr[i].len,tr[i].sum))%=p,k-=tr[i].sum;        else (ans*=qmul(tr[i].len,k))%=p,k=0;        if(!k)return ans;    }    return -1;}int main(){    //freopen("a.in","r",stdin);    scanf("%d%lld %s",&n,&k,s+1);    init();    rep(i,1,n)add(i);    //rep(i,1,node)printf("%d\n",tr[i].len);    per(i,node,1)tr[tr[i].suf].sum+=tr[i].sum;    sort(tr+1,tr+1+node);    printf("%d\n",solve());    return 0;}

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