题目链接：MAXMATCH - Maximum Self-Matching
Description
You're given a string s consisting of letters 'a', 'b' and 'c'.
The matching function \(m_s( i )\) is defined as the number of matching characters of s and its i-shift. In other words, \(m_s( i )\) is the number of characters that are matched when you align the 0-th character of s with the i-th character of its copy.
You are asked to compute the maximum of \(m_s( i )\) for all i ( 1 <= i <= |s| ). To make it a bit harder, you should also output all the optimal i's in increasing order.
Input
The first and only line of input contains the string s. \(2 \le |s| \le 10^5\).
Output
The first line of output contains the maximal \(m_s( i )\) over all i.
The second line of output contains all the i's for which \(m_s( i )\) reaches maximum.
Example
Input:
caccacaa
Output:
4
3
Explanation:
caccacaa
caccacaa
The bold characters indicate the ones that match when shift = 3.Solution
题意
给定一个字符串 \(s\) (下标从 \(0\) 开始，只包含 'a', 'b', 'c')，让 \(s\) 与 \(s\) 匹配，下标从 \(1\) 移动到 \(|s|\)，每次匹配时的相同的字符个数记为 \(m_s( i )\)，求 \(m_s( i )\) 的最大值，以及最大值所匹配的所有位置。
比如 ababa
ababa
ababa
ababa
ababa
ababa\(m_s( i )\) 分别为 \(0, 3, 0, 1\)，最大值为 \(3\)。
思路
FFT
字符串匹配问题。
设模式串为 \(p\)，目标串为 \(t\)，\(f[k]\) 为模式串从目标串第 \(k\) 位开始匹配的结果。
对 \(a\)，\(b\)，\(c\) 分开求。
首先判断 \(a\) 的情况，将字符串转化为 01 串，比如 ababa 转为 10101。
那么
\[f[k] = \sum_{i=0}^{|s|-k-1} p[i] \cdot t[k + i]\]
\[f[k] = \sum_{i=k}^{|s|-1} p[i - k] \cdot t[i]\]
将模式串倒置，有
\[f[k] = \sum_{i=k}^{|s|-1} p[|s| - 1 - i + k] \cdot t[i]\]
令 \(j = |s| - 1 - i + k\)，有
\[f[k] = \sum_{i+j=|s|-1+k} p[j] \cdot t[i]\]
用 \(FFT\) 求一下卷积即可。
对于 \(b\) 和 \(c\) 的求法相同。
Code
#include
using namespace std;
const double PI = acos(-1);
const double eps = 1e-8;
typedef complex Complex;
const int maxn = 2e6 + 10;
Complex p[maxn], t[maxn];
Complex a[maxn], b[maxn], c[maxn];
int ans[maxn];
string str;
int n;
int bit = 2, rev[maxn];
void get_rev(){
memset(rev, 0, sizeof(rev));
while(bit <= n + n) bit <<= 1;
for(int i = 0; i < bit; ++i) {
rev[i] = (rev[i >> 1] >> 1) | (bit >> 1) * (i & 1);
}
}
void FFT(Complex *arr, int op) {
for(int i = 0; i < bit; ++i) {
if(i < rev[i]) swap(arr[i], arr[rev[i]]);
}
for(int mid = 1; mid < bit; mid <<= 1) {
Complex wn = Complex(cos(PI / mid), op * sin(PI / mid));
for(int j = 0; j < bit; j += mid<<1) {
Complex w(1, 0);
for(int k = 0; k < mid; ++k, w = w * wn) {
Complex x = arr[j + k], y = w * arr[j + k + mid];
arr[j + k] = x + y, arr[j + k + mid] = x - y;
}
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> str;
n = str.size();
for(int i = 0; i < n; ++i) {
p[n - i - 1] = str[i] == 'a' ? 1 : 0;
t[i] = p[n - i - 1];
}
get_rev();
FFT(p, 1); FFT(t, 1);
for(int i = 0; i < bit; ++i) {
a[i] = p[i] * t[i];
}
FFT(a, -1);
for(int i = 0; i < n; ++i) {
p[n - i - 1] = str[i] == 'b' ? 1 : 0;
t[i] = p[n - i - 1];
}
for(int i = n; i < bit; ++i) {
p[i] = 0;
t[i] = 0;
}
FFT(p, 1); FFT(t, 1);
for(int i = 0; i < bit; ++i) {
b[i] = p[i] * t[i];
}
FFT(b, -1);
for(int i = 0; i < n; ++i) {
p[n - i - 1] = str[i] == 'c' ? 1 : 0;
t[i] = p[n - i - 1];
}
for(int i = n; i < bit; ++i) {
p[i] = 0;
t[i] = 0;
}
FFT(p, 1); FFT(t, 1);
for(int i = 0; i < bit; ++i) {
c[i] = p[i] * t[i];
}
FFT(c, -1);
int maxa = 0;
for(int i = 1; i < n; ++i) {
ans[i] = (int)(a[n - 1 + i].real() / bit + 0.5) + (int)(b[n - 1 + i].real() / bit + 0.5) + (int)(c[n - 1 + i].real() / bit + 0.5);
maxa = max(maxa, ans[i]);
}
vector pos;
for(int i = 1; i < n; ++i) {
if(ans[i] == maxa) {
pos.push_back(i);
}
}
cout << maxa << endl;
for(int i = 0; i < pos.size(); ++i) {
cout << pos[i] << " ";
}
cout << endl;
return 0;
}