## ZOJ 2619 Generator （概率、AC自动机、高斯消元）

2018年01月12日 ⁄ 综合 ⁄ 共 2759字 ⁄ 字号 评论关闭

# Generator

F[i] = 1 + 1 / n *
Σ
F[Ci,j] （其中Ci,j 表示在i状态末尾添加第j个字母之后，得到的新状态。
1. 上面可以得到一个方程组，解的的F[0]即为答案（PS 这道题用double解高斯消元过不了，所以要用解整数的高斯消元）
2. Ci,j 的实质是求给定字符串的第i个前缀添加字母后变成哪一个前缀，可以用AC自动机来预处理。

```/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
#define eps 1e-6
typedef long long ll;
typedef unsigned long long ULL;
using namespace std;
const int maxn = 15;
long long a[maxn][maxn] = {0}, ans[maxn] = {0};
bool l[maxn];
int n;
inline int solve(long long a[][maxn], bool l[], long long ans[], const int& n) {
int res = 0, r = 0;
for (int i = 0; i < n; i++) l[i] = false;
for (int i = 0; i < n; i++) {
for (int j = r; j < n; j++)
if (a[j][i] != 0) {
for (int k = i; k <= n; k++) swap(a[j][k], a[r][k]);
break;
}
if (a[r][i] == 0) {
res++;
continue;
}
for (int j = 0; j < n; j++)
if (j != r && a[j][i] != 0) {
if (a[j][i] < 0) {
for (int k = 0; k <= n; k++) a[j][k] *= -1;
}
long long gcd = __gcd(a[r][i], a[j][i]);
long long lcm = a[r][i] / gcd * a[j][i];
long long jmul = lcm / a[j][i];
long long imul = lcm / a[r][i];
for (int k = 0; k <= n; k++) {
a[j][k] = a[j][k] * jmul - a[r][k] * imul;
}
}
l[i] = true, r++;
}
for (int i = r; i < n; i++) if (a[i][n] != 0) return -1;
for (int i = 0; i < n; i++)
if (l[i])
for (int j = 0; j < n; j++)
if (a[j][i] != 0)
ans[i] = a[j][n] / a[j][i];

return res;
}

const int maxnode = 15 * 26;
int charset;
struct ACAutomaton {
int ch[maxnode][26];
int fail[maxnode];
int Q[maxnode];
int val[maxnode];
int sz;
int ID[128];
void init() {
fail[0] = 0;
for (int i = 0; i < 26; i++) ID[i + 'A'] = i;
}
void reset() {
sz = 1;
memset(ch[0], 0, sizeof(ch[0]));
}
void Insert(char* s, int key) {
int u = 0;
for (; *s; s++) {
int c = ID[*s];
if (!ch[u][c]) {
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = key;
}
void Construct () {
int *s = Q, *e = Q;
for (int i = 0; i < charset; i++) {
if (ch[0][i]) {
*e++ = ch[0][i];
fail[ch[0][i]] = 0;
}
}
while(s != e) {
int u = *s++;
if (val[fail[u]]) val[u] = 1;
for (int i = 0; i < charset; i++) {
int &v = ch[u][i];
if (v) {
*e++ = v;
fail[v] = ch[fail[u]][i];
} else {
v = ch[fail[u]][i];
}
}
}
}

void work() {
double t = 1.0 / charset;
//cout<<charset<<endl;
for (int i = 0; i < sz - 1; i++) {
a[i][i] += charset, a[i][sz - 1] += charset;
for (int j = 0; j < charset; j++) {
if (!val[ch[i][j]]) {
//cout<<i<<" "<<ch[i][j]<<endl;
a[i][ch[i][j]] -= 1;
}
}
}
}
} AC;

int main () {
int n, t, cas = 1;
char str[15];
scanf("%d", &t);
AC.init();
while(t--) {
scanf("%d%s", &n, str);
AC.reset();
charset = n;
AC.Insert(str, 1);
memset(a, 0, sizeof(a));
AC.Construct(); //得到状态转移
AC.work(); //构造方程
solve(a, l, ans, AC.sz - 1);
if (cas != 1) putchar('\n');
printf("Case %d:\n%lld\n", cas++, ans[0]);
}
return 0;
}
```