最小表示法+hash hdu2609 How many

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题意：有n个串，每个串是一个环，如果通过移动环之后能一样，认为是同一种串。问有多少种串。

思路：很明显我们首先要把认为是一样的串变成一样的，比如找到这个串的最小表示法，所以这道题的目的就是为了求最小表示法，然后用hash搞一搞排序然后去重，或者是直接插入到set里面取size都是可以的

KMP(复杂度O(n))来搞最小表示法暂时还不会，先挖个坑，我这里还有一种通过二分+hash求lcp来求最小表示法的思路，复杂度O(nlogn)，先凑合着用把

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout << "[" << x << "]"#define FIN freopen("input.txt", "r", stdin)#define FOUT freopen("output.txt", "w+", stdout)using namespace std;typedef long long LL;typedef pair<int, int> PII;typedef unsigned long long ULL;const int MX = 3e2 + 5;const int INF = 0x3f3f3f3f;const int seed = 131;ULL fac[MX], pre[MX], S[10005];char A[MX];void presolve() {    fac[0] = 1;    for(int i = 1; i < MX; i++) {        fac[i] = fac[i - 1] * seed;    }}bool check(int a, int b, int l) {    ULL left = pre[a + l - 1] - pre[a - 1] * fac[l];    ULL right = pre[b + l - 1] - pre[b - 1] * fac[l];    return left == right;}int lcp(int n, int a, int b) {    int l = 0, r = n, m;    while(l <= r) {        m = (l + r) >> 1;        if(check(a, b, m)) l = m + 1;        else r = m - 1;    }    return l - 1;}ULL change() {    int n = strlen(A + 1);    for(int i = 1; i <= n; i++) {        A[n + i] = A[i];    }    pre[0] = 0;    for(int i = 1; i <= 2 * n; i++) {        pre[i] = pre[i - 1] * seed + A[i];    }    int ans = 1, m;    for(int i = 1; i <= n; i++) {        int l = lcp(n, ans, i);        if(A[i + l] < A[ans + l]) ans = i;    }    ULL ret = 0;    for(int i = ans, j = 1; j <= n; i++, j++) {        ret = ret * seed + A[i];    }    return ret;}int main() {    int n; //FIN;    presolve();    while(~scanf("%d", &n)) {        for(int i = 1; i <= n; i++) {            scanf("%s", A + 1);            S[i] = change();        }        sort(S + 1, S + 1 + n);        int sz = unique(S + 1, S + 1 + n) - S - 1;        printf("%d\n", sz);    }    return 0;}

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