AC自动机+dp+大数 poj1625

传送门：点击打开链接

题意：告诉你有哪些字符可以用，然后再告诉你哪些单词不能出现，要你求长度为m的字符串只由给你的字符组成，但是不能出现那些单词的种类数。

思路：..一分析就发现，，貌似爆long long了，，醉了。。总的思路和那个DNA的思路是一样的，用AC自动机完成了矩阵的转移，很逆天的压缩了状态。。除了大数其他和那题基本一样的可以去看看那题..

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#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;

/*MX为总长度*/
const int MX = 2e2 + 5;
const int P = 128;

int G[MX][MX];

struct AC_machine {
    int rear,root,ID[MX],s;
    int Next[MX][P],Fail[MX],End[MX];

    void Init(char *A) {
        rear = 0;
        s = strlen(A);
        for(int i = 0; i < s; i++) {
            ID[A[i]] = i;
        }
        root = New();
    }

    int New() {
        End[rear] = 0;
        for(int i = 0; i < s; i++) {
            Next[rear][i] = -1;
        }
        return rear++;
    }

    void Add(char *A) {
        int now = root,n = strlen(A);
        for(int i = 0; i < n; i++) {
            int id = ID[A[i]];
            if(Next[now][id] == -1) {
                Next[now][id] = New();
            }
            now = Next[now][id];
        }
        End[now] = 1;
    }

    void Build() {
        queue<int>Q;
        Fail[root] = root;
        for(int i = 0; i < s; 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 u = Q.front(); Q.pop();

            if(End[Fail[u]]) End[u] = 1;
            for(int i = 0; i < s; i++) {
                if(Next[u][i] == -1) {
                    Next[u][i] = Next[Fail[u]][i];
                } else {
                    Fail[Next[u][i]] = Next[Fail[u]][i];
                    Q.push(Next[u][i]);
                }
            }
        }
    }

    int Query() {
        memset(G,sizeof(G));
        for(int i = 0; i < rear; i++) {
            for(int j = 0; j < s; j++) {
                if(End[Next[i][j]] == 0) {
                    G[i][Next[i][j]]++;
                }
            }
        }
        return rear;
    }
} AC;

struct BigInteger {
    int A[30];
    enum {MOD = 10000};
    BigInteger() {
        memset(A,sizeof(A));
        A[0] = 1;
    }
    void set(int x) {
        memset(A,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;
    }
};

char word[MX];
BigInteger dp[2][MX];

int main() {
    int n,m,p; //FIN;
    while(~scanf("%d%d%d",&n,&m,&p)) {
        memset(dp,sizeof(dp));
        scanf("%s",word);
        AC.Init(word);

        for(int i = 1; i <= p; i++) {
            scanf("%s",word);
            AC.Add(word);
        }
        AC.Build();
        int s = AC.Query();

        int cur = 0,nxt = 1;
        dp[cur][0].set(1);
        for(int i = 1; i <= m; i++) {
            for(int j = 0; j < s; j++) {
                dp[nxt][j].set(0);
            }

            for(int j = 0; j < s; j++) {
                for(int k = 0; k < s; k++) {
                    if(G[j][k]) {
                        BigInteger temp;
                        temp.set(G[j][k]);
                        dp[nxt][k] = dp[nxt][k] + dp[cur][j] * temp;
                    }
                }
            }
            swap(cur,nxt);
        }

        BigInteger ans;
        for(int i = 0; i < s; i++) {
            ans = ans + dp[cur][i];
        }
        ans.print();
    }
    return 0;
}