CF-528D Fuzzy Search(FFT字符串匹配)

Fuzzy Search

题意：

给定一个模式串和目标串按下图方式匹配，错开位置不多于k

解题思路：

总共只有\(A C G T\)四个字符，那么我们可以按照各个字符进行匹配，比如按照\(A\)进行匹配时,当\(k=1\)时，我们将目标串

\(ACAT\)化作

\(1~0~1~0\)

模式串

\(AGCAATTCAT\)化作

\(1~1~1~1~1~1~0~1~1~1\)

同样是反置目标串

可以得到以x为匹配终点的位置的匹配函数\(p(X)=\sum_{i+j=x}A(i)B(j)\)

如此进行4次FFT，最后如果目标位置贡献等于目标串长度，则说明匹配成功

#include <bits/stdc++.h>
using namespace std;
/*    freopen("k.in", "r", stdin);
    freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e6 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
template <class T>
inline void in(T &x)
{
    static char ch;
    static bool neg;
    for (ch = neg = 0; ch < '0' || '9' < ch; neg |= ch == '-', ch = getchar())
        ;
    for (x = 0; '0' <= ch && ch <= '9'; (x *= 10) += ch - '0', ch = getchar())
        ;
    x = neg ? -x : x;
}

struct Complex
{
    double x, y;
    Complex(double xx = 0, double yy = 0) { x = xx, y = yy; }
} a[MAXN], b[MAXN], c[MAXN], ans[MAXN];
Complex operator+(Complex a, Complex b) { return Complex(a.x + b.x, a.y + b.y); }
Complex operator-(Complex a, Complex b) { return Complex(a.x - b.x, a.y - b.y); }
Complex operator*(Complex a, Complex b) { return Complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } //不懂的看复数的运算那部分
int N, M;
int l, r[MAXN];
int limit = 1;
void FFT(Complex *A, int type)
{
    for (int i = 0; i < limit; i++)
        if (i < r[i])
            swap(A[i], A[r[i]]); //求出要迭代的序列
    for (int mid = 1; mid < limit; mid <<= 1)
    {                                                    //待合并区间的长度的一半
        Complex Wn(cos(pi / mid), type * sin(pi / mid)); //单位根
        for (int R = mid << 1, j = 0; j < limit; j += R)
        {                    //R是区间的长度，j表示前已经到哪个位置了
            Complex w(1, 0); //幂
            for (int k = 0; k < mid; k++, w = w * Wn)
            {                                                 //枚举左半部分
                Complex x = A[j + k], y = w * A[j + mid + k]; //蝴蝶效应
                A[j + k] = x + y;
                A[j + mid + k] = x - y;
            }
        }
    }
    /*if (type == -1)
        for (int i = 0; i < limit; ++i)
            a[i].x /= limit;//我们推过的公式里面有一个1/n这一项*/
}
char s[MAXN], t[MAXN];
void init(int N, int M)
{
    while (limit <= N + M)
        limit <<= 1, l++;
    for (int i = 0; i < limit; i++)
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
}
int change(char str)
{
    if (str == 'A')
        return 1;
    else if (str == 'T')
        return 2;
    else if (str == 'G')
        return 3;
    else
        return 4;
}
int pre[MAXN], cnt;
int main()
{
    int n, m, k;
    scanf("%d%d%d %s %s", &n, &m, &k, s, t);
    reverse(t, t + m);
    init(n, m);
    for (int ca = 1; ca <= 4; ca++)
    {
        cnt = -1;
        memset(pre, 0, sizeof(pre));
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for (int i = 0; i < n; i++)
        {
            if (change(s[i]) == ca)
                pre[++cnt] = i;
            a[i].x = change(s[i]) == ca ? 1 : 0, a[i].y = 0;
        }
        for (int i = 0; i < m; i++)
            b[i].x = change(t[i]) == ca ? 1 : 0, b[i].y = 0;
        int now = -1;
        for (int i = 0; i <= cnt; i++)
        {
            int L = max(pre[i] - k, 0);
            int R = min(pre[i] + k, n - 1);
            if (now > R)
                continue;
            now = max(L, now);
            for (; now <= R; now++)
                a[now].x = 1;
            now--;
        }
        FFT(a, 1);
        FFT(b, 1);
        for (int i = 0; i < limit; i++)
            a[i] = b[i] * a[i];
        FFT(a, -1);
        for (int i = 0; i < limit; i++)
            c[i] = c[i] + a[i];
    }
    int ans = 0;
    for (int i = m - 1; i < limit; i++)
        if (int(c[i].x / limit + 0.5) == m)
            ans++;
    printf("%d\n", ans);
    return 0;
}