Generator

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2619

题意：给定一个数N，代表可以选前N个字母。然后给定一个仅有前N个字母组成的字符串，问从空串开始构造，每次可以在已有基础上从前N个字母中挑选一个加在后面，问构造的字符串的长度期望是多少？

思路：如果给定的串长度为L，那么对于构造的串，对应的状态就是0到L之间的数。如果为L即为构造成功。记F[i] 为从i状态到达L状态期望的步数，那么对于0到L-1中的每个i，可以列出一个状态转移：

F[i] = 1 + 1 / n *
ΣF[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。

代码：

/*
ID: [email protected]
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;
}

ZOJ 2619 Generator （概率、AC自动机、高斯消元）,布布扣,bubuko.com