CF 954I - Yet Another String Matching Problem FFT 字符串

题意：

给你两个字符串S,T。

对于每个T长度的S子串，你需要求出他们的“距离”。

你每次可以将一个字符变成另一个字符，对于两个串都变，距离就是变化次数最小值使得两串相同。

字符集a - f。

题解：

如果长度相同，和CF另一题一模一样，只要把同一位置的两个字符连边，然后并查集一下就可以了。

现在还是一样，只是多了FFT，我们枚举转换字符i, j， 6 * 6 = 36种，然后把S中所有i字符变成1，其他的为0， T中所有j字符变成1，其他的为0。然后跑FFT，res[i]>0代表有这种边。

代码：

#include <bits/stdc++.h>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 3e5 + 5;char s[MAXN], t[MAXN];int par[MAXN][7];int Find (int x, int i) {return x == par[i][x] ? x : par[i][x] = Find (par[i][x], i);
}int Union (int x, int y, int i) {int xx = Find (x, i), yy = Find (y, i);if (xx != yy) {par[i][xx] = yy;return 1;}return 0;
}//typedef complex<double> CP;
struct CP {double x, y;CP() {}CP (double x, double y) : x (x), y (y) {}inline double real() {return x;}inline CP operator * (const CP& r) const {return CP (x * r.x - y * r.y, x * r.y + y * r.x);}inline CP operator - (const CP& r) const {return CP (x - r.x, y - r.y);}inline CP operator + (const CP& r) const {return CP (x + r.x, y + r.y);}inline CP conj (const CP &r) {return CP (r.x, -r.y);}
};
CP a[MAXN], b[MAXN];
int r[MAXN], res[MAXN];
void fft_init (int nm, int k) {
// Rader 操作for (int i = 0; i < nm; ++i) r[i] = (r[i >> 1] >> 1) | ( (i & 1) << (k - 1) );
}
void fft (CP ax[], int nm, int op) {for (int i = 0; i < nm; ++i) if (i < r[i]) swap (ax[i], ax[r[i]]);for (int h = 2, m = 1; h <= nm; h <<= 1, m <<= 1) { // 枚举长度CP wn = CP (cos (op * 2 * PI / h), sin (op * 2 * PI / h) );for (int i = 0; i < nm; i += h) { // 枚举所有长度为 h 的区间CP w (1, 0); // 旋转因子for (int j = i; j < i + m; ++j, w = w * wn) { // 枚举角度CP t = w * ax[j + m]; // 蝴蝶操作ax[j + m] = ax[j] - t;ax[j] = ax[j] + t;}}}if (op == -1) for (int i = 0; i < nm; ++i) ax[i].x /= nm;
}
void trans (int ax[], int bx[], int n, int m) {int nm = 1, k = 0;while (nm < 2 * n || nm < 2 * m) nm <<= 1, ++k;for (int i = 0; i < n; ++i) a[i] = CP (ax[i], 0);for (int i = 0; i < m; ++i) b[i] = CP (bx[i], 0);for (int i = n; i < nm; ++i) a[i] = CP (0, 0);for (int i = m; i < nm; ++i) b[i] = CP (0, 0);fft_init (nm, k);fft (a, nm, 1);fft (b, nm, 1);for (int i = 0; i < nm; ++i) a[i] = a[i] * b[i];fft (a, nm, -1);nm = n + m - 1;for (int i = 0; i < nm; ++i) res[i] = (int) (a[i].real() + 0.5);
}int ax[MAXN], bx[MAXN];
int ans[MAXN];int main() {
#ifdef LOCALfreopen ("input.txt", "r", stdin);
#endifscanf ("%s %s", s + 1, t + 1);int n = strlen (s + 1), m = strlen (t + 1);reverse (t + 1, t + 1 + m);for (int i = 1; i <= n; i++) for (int j = 0; j < 7; j++) par[i][j] = j;for (int i = 0; i < 6; i++) {for (int j = 0; j < 6; j++) {for (int k = 1; k <= n; k++) ax[k] = s[k] == i + 'a';for (int k = 1; k <= m; k++) bx[k] = t[k] == j + 'a';trans (ax, bx, n + 1, m + 1);for (int k = m + 1; k <= n + 1; k++) if (res[k]) ans[k - m] += Union (i, j, k - m);}}for (int i = 1; i <= n - m + 1; i++) printf ("%d ", ans[i]);return 0;
}

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CF 954I - Yet Another String Matching Problem FFT 字符串

题意：

题解：

代码：

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