4084: [Sdoi2015]bigyration hash+map

60分暴力。。BZOJ卡过去了。
我们先确定一个长串，那么总串的一半就确定了，然后把短串hash后放入map，枚举长串的断点放进去匹配，如果匹配成功就统计答案。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#define base 233
#define U unsigned int
#define N 4000005
using namespace std;
int total_s,total_t,len_s,len_t;
int ans,len;
U T[N],h[N];
map<U,int> mp1,mp2;
string str_s[N],str_t[N];
char s[N];
inline U get_hash(int l,int r)
{
    return h[r]-h[l-1]*T[r-l+1];
}
int main()
{
    scanf("%d%d%d%d",&total_s,&total_t,&len_s,&len_t);
    len=len_s+len_t>>1; 
    if (len_s<len_t) swap(total_s,total_t),swap(len_s,len_t);
    for (int i=1;i<=total_s;i++) {scanf("%s",s); str_s[i]=string(s);}
    for (int i=1;i<=total_t;i++) {scanf("%s",s); str_t[i]=string(s);}
    for (int i=1;i<=total_t;i++)
    {
        U hash=0;
        for (int j=1;j<=len_t;j++)
            hash=hash*base+str_t[i][j-1]-'a';
        mp1[hash]++;
    }
    T[0]=1;
    for (int i=1;i<=len_s;i++) T[i]=T[i-1]*base;
    for (int i=1;i<=total_s;i++)
    {
        mp2.clear();
        for (int j=1;j<=len_s;j++)
            h[j]=h[j-1]*base+str_s[i][j-1]-'a';
        for (int j=1;j<=len;j++)
        {
            U hash=get_hash(1,j-1)+get_hash(j,len)*T[j-1];
            if (mp2[hash]) continue;
            mp2[hash]=1;
            ans+=mp1[hash-get_hash(len+1,len_s)*T[len_t]];
        }
    }
    cout << ans << endl;
    return 0;
}