NOIP 模拟 $25\; \rm string$

题解 \(by\;zj\varphi\)

考虑对于母串的每个字符，它在匹配串中有多少前缀，多少后缀。

设 \(f_i\) 表示 \(i\) 位置匹配上的前缀，\(g_i\) 为后缀，那么答案为 \(\sum_{i=1}^{len}f_i×g_i\)

那么如何求出 \(f_i\) 和 \(g_i\)，考虑二分，求出一个最长的前缀，后缀。

在初始化时，将所有后缀记录上它的子后缀，前缀同理，用 \(trie\) 树就行，记得用 unordered_map

Code:

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf,OPUT[100];
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1<<21,1,stdin),p1==p2)?(-1):*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=getchar();
        while(!isdigit(ch)) {if (ch=='-') f=0;ch=getchar();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
        x=f?x:-x;
    }
    template<typename T>inline void print(T x) {
        if (x<0) putchar('-'),x=-x;
        if (!x) return putchar('0'),(void)putchar('\n');
        ri cnt(0);
        while(x) OPUT[p(cnt)]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
        return (void)putchar('\n');
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    typedef unsigned long long ull;
    typedef long long ll;
    static const int N=1e5+7,P=131,L=2e5+7;
    int len1,len,n;
    ll ans;
    char s[N],s1[N];
    ull p[N],h[N];
    unordered_map<ull,int> mp1,mp2;
    struct Trie{
        #define Son(x,p) T[x].ch[p]
        struct trie{int ch[26],nm;}T[L];
        int tot;
        Trie(){tot=1;}
        inline void insert() {
            ri cur=1;
            for (ri i(1);i<=len;p(i)) {
                ri ch=s[i]-'a';
                if (!Son(cur,ch)) Son(cur,ch)=p(tot);
                cur=Son(cur,ch);
                p(T[cur].nm);
            }
        }
        void dfs1(int nw,ull h) {
            if (nw!=1&&T[nw].nm) mp1[h]=T[nw].nm;
            for (ri i(0);i<26;p(i))
                if (Son(nw,i)) {
                    T[Son(nw,i)].nm+=T[nw].nm;
                    dfs1(Son(nw,i),(ull)(i+1)+h*P);
                }
        }
        void dfs2(int nw,ull h,int dep) {
            if (nw!=1&&T[nw].nm) mp2[h]=T[nw].nm;
            for (ri i(0);i<26;p(i))
                if (Son(nw,i)) {
                    T[Son(nw,i)].nm+=T[nw].nm;
                    dfs2(Son(nw,i),(ull)(i+1)*p[dep]+h,dep+1);
                }
        }
    }T1,T2;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        p[0]=1;
        for (ri i(1);i<=N-7;p(i)) p[i]=p[i-1]*P;
        scanf("%s",s1+1);
        len1=strlen(s1+1);
        for (ri i(1);i<=len1;p(i)) h[i]=h[i-1]*P+(ull)(s1[i]-'a'+1);
        read(n);
        ull k=-1;
        for (ri i(1);i<=n;p(i)) {
            scanf("%s",s+1);
            len=strlen(s+1);
            T1.insert();
            reverse(s+1,s+len+1);
            T2.insert();
        }
        T1.dfs1(1,0),T2.dfs2(1,0,0);
        for (ri i(1);i<len1;p(i)) {
            ri l(1),r(i),res(-1),tmp1(0),tmp2(0);
            while(l<=r) {
                int mid(l+r>>1);
                if (mp2.find(h[i]-h[i-mid]*p[mid])!=mp2.end()) l=mid+1,res=mid;
                else r=mid-1;
            }
            if (res!=-1) tmp2=mp2[h[i]-h[i-res]*p[res]];
            l=1,r=len1-i,res=-1;
            while(l<=r) {
                int mid(l+r>>1);
                if (mp1.find(h[i+mid]-h[i]*p[mid])!=mp1.end()) l=mid+1,res=mid;
                else r=mid-1;
            }
            if (res!=-1) tmp1=mp1[h[i+res]-h[i]*p[res]];
            ans+=1ll*tmp1*tmp2;
        }
        print(ans);
        return 0;
    }
}
int main() {return nanfeng::main();}