BZOJ 4084: [Sdoi2015]bigyration

Description

Input

Output

Sample Input

4 4 7 3

vijosvi

josvivi

vijosos

ijosvsv

jos

vij

ijo

jos

Sample Output

6

HINT

Solution

短的字符串先Hash一下,统计出现次数。然后把长的字符串前一半拿出来复制一遍接在后面,跑Hash。然后枚举长的字符串后一半开始位置,如果匹配,答案加上后面字符出现次数。

配上张图来讲一下:

我们假设长字符串为S,那么绿色就是S的后半部分。黑色是S的前半部分,深灰色的是我们复制的。浅灰色的是S的长度。然后我们可以发现绿色的成功匹配了复制后的字符串的某一段,那么我们就找出他剩下的那部分(蓝色)的哈希,然后看看它出现了几次。

哈希大法好!!!

Code

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#include <cstdio>
#include <algorithm>
#include <map>
#include <string>
using namespace std;
typedef unsigned long long ULL;
const int INF = 1009999999;
char str[4000005];
int ts, tt, n, m;
string wa[4000005], wb[4000005];
ULL H[4000005], mul[4000005];
inline ULL GetHash(int l, int r) { --l; return H[r] - H[l] * mul[r - l]; }
map<ULL, int> M;
int main() {
    scanf("%d%d%d%d", &ts, &tt, &n, &m);
    int HSK = (n + m) >> 1;
    string *S = wa, *T = wb;
    for (int i = 0; i < ts; ++i) {
        scanf("%s", str);
        S[i] = string(str);
    }
    for (int i = 0; i < tt; ++i) {
        scanf("%s", str);
        T[i] = string(str);
    }
    if (n < m) {
        swap(n, m);
        swap(S, T);
        swap(ts, tt);
    }
    mul[0] = 1;
    for (int i = 1; i <= n; ++i)
        mul[i] = mul[i-1] * 29;
    ULL has, Mag, Ans = 0;
    for (int i = 0; i < tt; ++i) {
        has = 0;
        for (int j = 0; j < m; ++j)
            has = has * 29 + ULL(T[i][j] - 'a' + 1);
        M[has] ++;
    }
    for (int i = 0; i < ts; ++i) {
        has = 0;
        for (int j = 0; j < HSK; ++j)
            H[j + 1] = H[j] * 29 + ULL(S[i][j] - 'a' + 1);
        for (int j = HSK; j < (HSK << 1); ++j)
            H[j + 1] = H[j] * 29 + ULL(S[i][j - HSK] - 'a' + 1);
        for (int j = HSK; j < n; ++j)
            has = has * 29 + ULL(S[i][j] - 'a' + 1);
        for (int j = 1; j <= HSK; ++j) {
            if (GetHash(j, j + n - HSK - 1) == has) {
                Mag = GetHash(j + n - HSK, j + HSK - 1);
                Ans += M[Mag];
            }
        }
    }
    printf("%llu", Ans);
    return 0;
}