poj 1625 （AC自动机好模版，大数好模版）

题目

给n个字母，构成长度为m的串，总共有n^m种。给p个字符串，问n^m种字符串中不包含（不是子串）这p个字符串的个数。

将p个不能包含的字符串建立AC自动机，每个结点用val值来标记以当前节点为后缀的字符串是否包含非法字符串（p个字符串中的任何一个）。

状态转移方程：f(i, j)  += f(i-1, k)  

f(i, j)表示长度为i的字符串，结尾为字符j，方程j和k的关系可以从自动机中失配关系直接获得（j是k的后继结点）。

 

总之感觉是好东西，快存下来

 

大数模版：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

struct BigInteger{
    int A[25];
    enum{MOD = 10000};
    BigInteger(){memset(A, 0, sizeof(A)); A[0]=1;}
    void set(int x){memset(A, 0, sizeof(A)); A[0]=1; A[1]=x;}
    void print(){
        printf("%d", A[A[0]]);
        for (int i=A[0]-1; i>0; i--){
            if (A[i]==0){printf("0000"); continue;}
            for (int k=10; k*A[i]<MOD; k*=10) printf("0");
            printf("%d", A[i]);
        }
        printf("\n");
    }
    int& operator [] (int p) {return A[p];}
    const int& operator [] (int p) const {return A[p];}
    BigInteger operator + (const BigInteger& B){
        BigInteger C;
        C[0]=max(A[0], B[0]);
        for (int i=1; i<=C[0]; i++)
            C[i]+=A[i]+B[i], C[i+1]+=C[i]/MOD, C[i]%=MOD;
        if (C[C[0]+1] > 0) C[0]++;
        return C;
    }
    BigInteger operator * (const BigInteger& B){
        BigInteger C;
        C[0]=A[0]+B[0];
        for (int i=1; i<=A[0]; i++)
            for (int j=1; j<=B[0]; j++){
                C[i+j-1]+=A[i]*B[j], C[i+j]+=C[i+j-1]/MOD, C[i+j-1]%=MOD;
            }
        if (C[C[0]] == 0) C[0]--;
        return C;
    }
};
int main() {
    BigInteger a, b;
    a.set(1); b.set(1);
    (a+b).print();

    return 0;
}

View Code

 

本题答案（内含AC自动机模版）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef unsigned char uchar;

struct AC_Automata {
    #define N 102
    int ch[N][55], val[N], last[N], f[N], sz;
    void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }

    int hash[256], M;
    void set_hash(int n, uchar s[]) {
        M = n; for (int i=0; i<n; i++) hash[s[i]] = i;
    }
    void insert(uchar s[], int v) {
        int u = 0;
        for (int i=0; s[i]; i++) {
            int c = hash[s[i]];
            if (!ch[u][c]) {
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] = v;  //标记当前串为非法的
    }
    void build() {
        queue<int> q;
        f[0] = 0;
        for (int c=0; c<M; c++) {
            int u = ch[0][c];
            if (u) { f[u] = last[u] = 0; q.push(u); }
        }
        while (!q.empty()) {
            int r = q.front(); q.pop();
            for (int c=0; c<M; c++) {
                int u = ch[r][c];
                val[r] = val[r] || val[f[r]];   //判断当前结点是否有非法后缀
                if (!u) {
                    ch[r][c] = ch[f[r]][c];
                    continue;
                }
                q.push(u);
                f[u] = ch[f[r]][c];
                last[u] = val[f[u]] ? f[u] : last[f[u]];
            }
        }
    }
} ac;
struct BigInteger{
    int A[25];
    enum{MOD = 10000};
    BigInteger(){memset(A, 0, sizeof(A)); A[0]=1;}
    void set(int x){memset(A, 0, sizeof(A)); A[0]=1; A[1]=x;}
    void print(){
        printf("%d", A[A[0]]);
        for (int i=A[0]-1; i>0; i--){
            if (A[i]==0){printf("0000"); continue;}
            for (int k=10; k*A[i]<MOD; k*=10) printf("0");
            printf("%d", A[i]);
        }
        printf("\n");
    }
    int& operator [] (int p) {return A[p];}
    const int& operator [] (int p) const {return A[p];}
    BigInteger operator + (const BigInteger& B){
        BigInteger C;
        C[0]=max(A[0], B[0]);
        for (int i=1; i<=C[0]; i++)
            C[i]+=A[i]+B[i], C[i+1]+=C[i]/MOD, C[i]%=MOD;
        if (C[C[0]+1] > 0) C[0]++;
        return C;
    }
    BigInteger operator * (const BigInteger& B){
        BigInteger C;
        C[0]=A[0]+B[0];
        for (int i=1; i<=A[0]; i++)
            for (int j=1; j<=B[0]; j++){
                C[i+j-1]+=A[i]*B[j], C[i+j]+=C[i+j-1]/MOD, C[i+j-1]%=MOD;
            }
        if (C[C[0]] == 0) C[0]--;
        return C;
    }
};
int n, m, p;
uchar s[55];

int main() {

    while (scanf("%d %d %d ", &n, &m, &p) == 3) {
        ac.clear();
        cin >> s; ac.set_hash(n, s);
        while (p--) {
            cin >> s; ac.insert(s, 1);
        }
        ac.build();

        BigInteger f[51][101];
        f[0][0].set(1);

        for (int i=1; i<=m; i++)
            for (int j=0; j<ac.sz; j++)
                for (int k=0; k<n; k++) {
                    int u = ac.ch[j][k];
                    if (!ac.val[u]) f[i][u] = f[i][u] + f[i-1][j];
                }
        BigInteger ans;
        for (int i=0; i<ac.sz; i++)
            if (!ac.val[i]) ans = ans + f[m][i];
        ans.print();
    }
    return 0;
}

View Code