P4173 残缺的字符串
Posted Jozky86
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题意：
有A，B两个串，每个串都有通配符，问A为模板串，对于 B 的每一个位置 i，从这个位置开始连续 m 个字符形成的子串是否可能与 A 串完全匹配？
题解：
我们定义两个字符串S，T的距离为：
dis(S,T)=
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\\sum_{i=1}^{n-1}(S_{i}-T_{i})^2*S_{i}*T_{i}
∑i=1n−1(Si−Ti)2∗Si∗Ti
当T中以i结尾的串与S能匹配的条件为：
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dis(S,T_{i-m+1,i})=0
dis(S,Ti−m+1,i)=0
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f_{i}=\\sum_{j=0}^{i}(S_{j}-T_{i-j})^2*S_{j}*T_{i-j}=\\sum_{j=0}^{i}S_{j}^{3}*T_{i-j}-2*\\sum_{j=0}^{i}S_{j}^{2}T_{i-j}^{2}+\\sum_{j=0}^{i}S_{j}*T_{i-j}^{3}
fi=∑j=0i(Sj−Ti−j)2∗Sj∗Ti−j=∑j=0iSj3∗Ti−j−2∗∑j=0iSj2Ti−j2+∑j=0iSj∗Ti−j3
我的板子
人傻了
这里贴的是别人的板子，开氧过了，不开80
代码：
// Problem: P4173 残缺的字符串
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4173
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// Data:2021-08-24 00:29:28
// By Jozky
#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
x= 0;
char c= getchar();
bool flag= 0;
while (c < '0' || c > '9')
flag|= (c == '-'), c= getchar();
while (c >= '0' && c <= '9')
x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
if (flag)
x= -x;
read(Ar...);
}
template <typename T> inline void write(T x)
{
if (x < 0) {
x= ~(x - 1);
putchar('-');
}
if (x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
startTime= clock();
freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
endTime= clock();
printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
#define MAXN 2000005
#define reg register
#define inl inline
#define db double
#define eps 1e-6
using namespace std;
const int Mod= 998244353;
const db Pi= acos(-1.0);
struct Complex
{
db x, y;
friend Complex operator+(const Complex& a, const Complex& b)
{
return ((Complex){a.x + b.x, a.y + b.y});
}
friend Complex operator-(const Complex& a, const Complex& b)
{
return ((Complex){a.x - b.x, a.y - b.y});
}
friend Complex operator*(const Complex& a, const Complex& b)
{
return ((Complex){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x});
}
friend Complex operator*(const Complex& a, const db& val)
{
return ((Complex){a.x * val, a.y * val});
}
} f[MAXN], g[MAXN], p[MAXN];
int n, m, lim= 1, maxn, rev[MAXN], a[MAXN], b[MAXN];
char S[MAXN], T[MAXN];
bool used[MAXN];
vector<int> v;
inl void FFT(reg Complex* A, reg int opt)
{
for (reg int i= 0; i < lim; i++)
if (i < rev[i])
swap(A[i], A[rev[i]]);
for (reg int mid= 1; mid < lim; mid<<= 1) {
reg Complex Wn= ((Complex){cos(Pi / (db)mid), (db)opt * sin(Pi / (db)mid)});
for (reg int j= 0; j < lim; j+= (mid << 1)) {
reg Complex W= ((Complex){1, 0});
for (reg int k= 0; k < mid; k++, W= W * Wn) {
reg Complex x= A[j + k], y= W * A[j + k + mid];
A[j + k]= x + y;
A[j + k + mid]= x - y;
}
}
}
}
int main()
{
scanf("%d %d", &m, &n);
scanf("%s", T + 1);
scanf("%s", S + 1);
for (reg int i= 1; i <= m; i++)
if (T[i] != '*')
a[i - 1]= T[i] - 'a' + 1;
for (reg int i= 1; i <= n; i++)
if (S[i] != '*')
b[i - 1]= S[i] - 'a' + 1;
while (lim <= (n + m)) {
lim<<= 1;
maxn++;
}
for (reg int i= 0; i < lim; i++)
rev[i]= ((rev[i >> 1] >> 1) | ((i & 1) << maxn - 1));
reverse(a, a + m);
for (reg int i= 0; i < m; i++)
f[i]= ((Complex){a[i] * a[i] * a[i], 0});
for (reg int i= 0; i < n; i++)
g[i]= ((Complex){b[i], 0});
FFT(f, 1);
FFT(g, 1);
for (reg int i= 0; i < lim; i++)
p[i]= p[i] + f[i] * g[i];
for (reg int i= 0; i < lim; i++)
f[i]= g[i]= ((Complex){0, 0});
for (reg int i= 0; i < m; i++)
f[i]= ((Complex){a[i] * a[i], 0});
for (reg int i= 0; i < n; i++)
g[i]= ((Complex){b[i] * b[i], 0});
FFT(f, 1);
FFT(g, 1);
for (reg int i= 0; i < lim; i++)
p[i]= p[i] - f[i] 以上是关于P4173 残缺的字符串的主要内容，如果未能解决你的问题，请参考以下文章