bzoj3670: [Noi2014]动物园

末蓝、 2021-12-09 03:43 291阅读 0赞

题意:求a\[1:i\]的2\*|border|<=i的num+1乘积  
题解:建kmp自动机(即next\[i\]指向i),由于某个点到根就是a\[1:i\]的border,然后树上二分即可

//#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>
    //#include <bits/extc++.h>
    #define fi first
    #define se second
    #define db double
    #define mp make_pair
    #define pb push_back
    #define mt make_tuple
    #define pi acos(-1.0)
    #define ll long long
    #define vi vector<int>
    #define mod 1000000007
    #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 ull unsigned long long
    //#define base 1000000000000000000
    #define fin freopen("a.txt","r",stdin)
    #define fout freopen("a.txt","w",stdout)
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
    inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
    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 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;
    //using namespace __gnu_pbds;
    
    const ull ba=233;
    const db eps=1e-5;
    const ll INF=0x3f3f3f3f3f3f3f3f;
    const int N=1000000+10,maxn=1000000+10,inf=0x3f3f3f3f;
    
    vi v[N];
    char s[N];
    int fail[N],st[N],top;
    void getfail()
    {
        fail[0]=-1;
        int k=-1,n=strlen(s);
        for(int i=0;i<=n;i++)v[i].clear();
        v[0].pb(1);
        for(int i=1;i<n;i++)
        {
            while(k>-1&&s[k+1]!=s[i])k=fail[k];
            if(s[k+1]==s[i])k++;
            fail[i]=k;
            v[fail[i]+1].pb(i+1);
        }
    }
    ll ans;
    void dfs(int u)
    {
        if(u)st[++top]=u;
        int l=0,r=top;
        while(l<r-1)
        {
            int m=(l+r)>>1;
            if(st[m]*2<=u)l=m;
            else r=m;
        }
    //    printf("%d %d\n",l,u);
        ans=ans*(l+1)%mod;
        for(int i=0;i<v[u].size();i++)dfs(v[u][i]);
        if(u)--top;
    }
    int main()
    {
        int T;scanf("%d",&T);
        while(T--)
        {
            scanf("%s",s);
            getfail();
            top=0;ans=1;dfs(0);
            printf("%lld\n",ans);
        }
        return 0;
    }
    /********************
    
    ********************/

转载于:https://www.cnblogs.com/acjiumeng/p/10792259.html