【BZOJ4503】两个串

【题目链接】

• BZOJ4503(权限题)

【前置技能】

• FFT/NTT

【题解】

• 字符串中出现了通配符，一般的字符串算法就失去效果了。
• 先忽略通配符的问题。得到一种算法：令每个位置$A_k=\displaystyle\sum_{i=0}^{LenT-1} (S_{i+k-LenT+1}-T_i)$,若$A_k=0$，则说明$T$与$S[k-LenT+1,k]$匹配。但发现这会出错，相减项有可能会正负抵消，如字符串ab和ba就会被认为是匹配的。得到了一种解决方法：将求和中的减法变成相减之后平方就可以避免抵消的问题了。
• 把T整个串翻转一下，然后把平方展开，发现这是卷积的形式。那么通配符的问题也很好想明白了，在式子上再乘上一个$T_i$，其中通配符的值为零即可。
• 最后的式子：$A_k=\displaystyle\sum_{i=0}^{k} (T_{k-i}*S_i^2)-2*\displaystyle\sum_{i=0}^{k} (T_{k-i}^2*S_i)+\displaystyle\sum_{i=0}^{k} T_{k-i}^3$。
• 时间复杂度$O(NlogN)$

【代码】

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL  long long
#define MAXN    262150
#define EPS 1e-3
using namespace std;
const double PI = acos(-1);
char s[MAXN], t[MAXN];
int n, LOG, ls, lt, rev[MAXN]; 
struct dot{double x, y;}ans[MAXN], a[MAXN], b[MAXN];
dot operator + (dot a, dot b){return (dot){a.x + b.x, a.y + b.y};}
dot operator - (dot a, dot b){return (dot){a.x - b.x, a.y - b.y};}
dot operator * (dot a, dot b){return (dot){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
dot operator / (dot a, double k){return (dot){a.x / k, a.y / k};}

template <typename T> void chkmin(T &x, T y){x = min(x, y);}
template <typename T> void chkmax(T &x, T y){x = max(x, y);}
template <typename T> void read(T &x){
    x = 0; int f = 1; char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
    x *= f;
}

void fft_init(){
    rev[0] = 0;
    for (int i = 1; i < n; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (LOG - 1));
}

void fft(dot *a, int opt){
    for (int i = 0; i < n; ++i)
        if (rev[i] < i) swap(a[rev[i]], a[i]);
    for (int len = 2; len <= n; len <<= 1){
        dot delta = (dot){cos(PI * 2 / len), sin(PI * 2 / len * opt)};
        for (int i = 0; i < n; i += len){
            dot cur = (dot){1, 0};
            for (int j = i, k = i + len / 2; k < i + len; ++j, ++k){
                dot tmp = a[j], tnp = a[k] * cur;
                a[j] = tmp + tnp;
                a[k] = tmp - tnp;
                cur = cur * delta;
            }
        }
    }
    if (opt == -1){
        for (int i = 0; i < n; ++i)
            a[i] = a[i] / n;
    }
}

bool sam(double x, double y){
    if (x > y - EPS && x < y + EPS) return 1;
    return 0;
}

int id(char ch){
    if (ch == '?' || ch == '\0') return 0;
    else return (ch - 'a' + 1);
}

int main(){
    scanf("%s", s);
    scanf("%s", t);
    ls = strlen(s); lt = strlen(t);
    reverse(t, t + lt);
    n = 1, LOG = 0;
    while (n <= ls + lt) n <<= 1, ++LOG;
    fft_init();
    for (int i = 0; i < n; ++i)
        a[i].x = id(t[i]), a[i].y = 0,
        b[i].x = id(s[i]), b[i].y = 0, b[i] = b[i] * b[i];
    for (int i = 0; i < n; ++i)
        ans[i] = a[i] * a[i] * a[i];
    for (int i = 1; i < n; ++i)
        ans[i] = ans[i] + ans[i - 1];

    fft(a, 1); fft(b, 1);
    for (int i = 0; i < n; ++i)
        a[i] = a[i] * b[i];
    fft(a, -1);
    for (int i = 0; i < n; ++i)
        ans[i] = ans[i] + a[i];

    for (int i = 0; i < n; ++i)
        a[i].x = id(t[i]), a[i].y = 0, a[i] = a[i] * a[i],
        b[i].x = id(s[i]), b[i].y = 0;
    fft(a, 1); fft(b, 1);
    for (int i = 0; i < n; ++i)
        a[i] = a[i] * b[i];
    fft(a, -1);
    for (int i = 0; i < n; ++i)
        ans[i] = ans[i] - (a[i] / 0.5);

    int cnt = 0;
    for (int i = lt - 1; i < ls; ++i)
        if (sam(ans[i].x, 0)) ++cnt;
    printf("%d\n", cnt);
    for (int i = lt - 1; i < ls; ++i)
        if (sam(ans[i].x, 0)) printf("%d\n", i - lt + 1);
    return 0;
}