【FFT】 Codeforces Round #296 (Div. 1) D - Fuzzy Search

先将t串反转，然后对每一个字母做多项式乘法。。。全加起来就是t串匹配的最大字母数。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 700005
#define maxm 1000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head

struct complex
{
	double r, i;
	complex(double r = 0, double i = 0) : r(r), i(i) {}
	complex operator + (const complex b) const {
		return complex(r + b.r, i + b.i);
	}
	complex operator - (const complex b) const {
		return complex(r - b.r, i - b.i);
	}
	complex operator * (const complex b) const {
		return complex(r * b.r - i * b.i, r * b.i + i * b.r);
	}
}a[maxn], b[maxn], res[maxn];

void fft(complex y[], int len, int on)
{
	for(int i = 1, j = len / 2; i < len-1; i++) {
		if(i < j) swap(y[i], y[j]);
		int k = len / 2;
		while(j >= k) {
			j -= k;
			k /= 2;
		}
		j += k;
	}
	for(int i = 2; i <= len; i <<= 1) {
		complex wn = complex(cos(-on * 2 * PI / i), sin(-on * 2 * PI / i));
		for(int j = 0; j < len; j += i) {
			complex w = complex(1.0, 0.0);
			for(int k = j; k < j + i / 2; k++) {
				complex u = y[k];
				complex t = w * y[k + i / 2];
				y[k] = u + t;
				y[k + i / 2] = u - t;
				w = w * wn;
			}
		}
	}
	if(on == -1) for(int i = 0; i < len; i++) y[i].r /= len;
}


char s[maxn];
char t[maxn];
int sum[maxn];
int n, m, kk;

void read()
{
	scanf("%d%d%d", &n, &m, &kk);
	scanf("%s%s", s, t);
}

void solve(char c, int len)
{
	memset(a, 0, sizeof a);
	memset(b, 0, sizeof b);
	memset(sum, 0, sizeof sum);
	for(int i = 0; i < n; i++) sum[i+1] = s[i] == c;
	for(int i = 1; i <= n; i++) sum[i] += sum[i-1];
	for(int i = 0; i < n; i++) {
		int mi = max(0, i - kk);
		int mx = min(n - 1, i + kk) + 1;
		if(sum[mx] - sum[mi]) a[i].r = 1.0;
	}
	for(int i = 0; i < m; i++) b[i].r = t[i] == c;
	fft(a, len, 1);
	fft(b, len, 1);
	for(int i = 0; i < len; i++) res[i] = res[i] + a[i] * b[i];
}

void work()
{
	reverse(t, t+m);
	int mx = max(n, m);
	int len = 1;
	while(len < 2 * mx) len <<= 1;
	solve('A', len);
	solve('C', len);
	solve('G', len);
	solve('T', len);
	fft(res, len, -1);
	int ans = 0;
	for(int i = m-1; i < n; i++) if((int)(res[i].r + 0.5) >= m) ans++;
	printf("%d\n", ans);
}

int main()
{
	read();
	work();
	
	return 0;
}