UVAlive 3490 AC自动机+（整数）高斯消元

题意：

题目链接：https://vjudge.net/problem/UVALive-3490
给出一个字符串S和前n个英文大写字母，组成一个字符串，当构造出子串S时停止，则该字符串期望长度是多少。

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思路：

AC自动机+高斯消元，与HDU 5955一个套路：http://blog.youkuaiyun.com/bahuia/article/details/53034010
这题坑的时直接用double的高斯消元误差太大，连样例都出不来，必须改成整形的高斯消元，学习别人的写法，存个板子。

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代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 100;

struct ACauto {
    int next[MAXN][26], fail[MAXN], end[MAXN];
    int root, sz;

    int newnode() {
        for (int i = 0; i < 26; i++)
            next[sz][i] = -1;
        end[sz++] = 0;
        return sz - 1;
    }

    void init() {
        sz = 0;
        root = newnode();
    }

    int idx(char c) {
        return c - 'A';
    }

    void insert(char *buf) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            int id = idx(buf[i]);
            if (next[now][id] == -1)
                next[now][id] = newnode();
            now = next[now][id];
        }
        end[now]++;
    }

    void build() {
        queue <int> Q;
        fail[root] = root;
        for (int i = 0; i < 26; i++) {
            if (next[root][i] == -1)
                next[root][i] = root;
            else {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        }
        while (!Q.empty()) {
            int now = Q.front(); Q.pop();
            for (int i = 0; i < 26; i++) {
                if (next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }

} ac;

int equ, var;
LL a[MAXN][MAXN], x[MAXN];

void Gauss()//表示用double版本的高斯消元求解,最后还原成整数的方法因为误差跪掉了....
{
    for(int i = 0; i < equ; i++)
    {
        int r = i;
        while(r < equ && !a[r][i]) r++;
        if(r != i)//找到第r列不是0的之后交换至r行
        {
            for(int j = 0; j < var; j++) swap(a[r][j], a[i][j]);
            swap(x[r], x[i]);
        }
        for(int k = i + 1; k < equ; k++)
            if(a[k][i])//将下面所有行的第i列变成0
            {
                LL tmp = a[k][i];//两组互相乘上对面的第i列的数作差即可
                for(int j = i; j < var; j++) a[k][j] = a[k][j] * a[i][i] - tmp*a[i][j];
                x[k] = x[k]*a[i][i] - tmp*x[i];
            }
    }
    //对剩下的三角矩阵递推求解
    for(int i = equ - 1; i >= 0; i--)
    {
        for(int j = i + 1; j < var; j++)
            x[i] -= x[j]*a[i][j];
        x[i] /= a[i][i];
    }
}


char str[MAXN];

int main() {
    //freopen("in.txt", "r", stdin);
    int T, cs = 0;
    scanf("%d", &T);
    while (T--) {
        int n;
        scanf("%d%s", &n, str);
        ac.init();
        ac.insert(str);
        ac.build();
        equ = var = ac.sz;
        memset(a, 0, sizeof(a));
        memset(x, 0, sizeof(x));
        //cout << ac.sz << endl;
        for (int u = 0; u < ac.sz - 1; u++) {
            a[u][u] = n;
            for (int id = 0; id < n; id++) {
                int v = ac.next[u][id];
                a[u][v] -= 1;
            }
            a[u][var] = n;
            x[u] = n;
        }
        a[equ - 1][var - 1] = 1;
        a[equ - 1][var] = 0;
        x[equ - 1] = 0;
        Gauss();
        printf("Case %d:\n%lld\n", ++cs, x[0]);
        if (T) puts("");
    }
    return 0;
}