bzoj5337 [TJOI2018]str（Hash+dp）

用hash处理字符串匹配，变成若干线段接力覆盖的问题，瞎dp一下就好了qaq
复杂度$O(m(len+n))$

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 10010
#define ull unsigned long long
#define k1 11113
#define mod 1000000007
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,m,a[2][N];
char s[N],t[N];
ull bin[N],hs[N];
inline ull calhs(int y,int x){
    return hs[y]-hs[x-1]*bin[y-x+1];
}
inline void inc(int &x,int y){x+=y;x%=mod;}
int main(){
//  freopen("str.in","r",stdin);
//  freopen("str.out","w",stdout);
    m=read();scanf("%s",s+1);n=strlen(s+1);
    bin[0]=1;int p=0;for(int i=0;i<=n;++i) a[0][i]=1;
    for(int i=1;i<=n;++i) bin[i]=bin[i-1]*k1,hs[i]=hs[i-1]*k1+s[i];
    for(int ii=1;ii<=m;++ii){
        int owo=read();
        while(owo--){
            scanf("%s",t+1);int len=strlen(t+1);ull hss=0;
            for(int i=1;i<=len;++i) hss=hss*k1+t[i];
            for(int i=1;i<=n;++i){
                if(i+len-1>n) break;
                if(calhs(i+len-1,i)!=hss) continue;
                inc(a[p^1][i+len-1],a[p][i-1]);
            }
        }memset(a[p],0,sizeof(a[p]));p^=1;
    }int ans=0;
    for(int i=1;i<=n;++i) inc(ans,a[p][i]);
    printf("%d\n",ans);
    return 0;
}