HDU 2243 考研路茫茫——单词情结 ( Trie图 && DP && 矩阵构造幂和 )

#include<string.h>
#include<stdio.h>
#include<iostream>
#include<queue>
#define ULL unsigned long long
using namespace std;

const int Max_Tot = 1e2 + 10;
const int Letter  = 26;
int maxn;///矩阵的大小
char S[11];

struct mat{ ULL m[111][111]; }unit, M;
mat operator * (mat a, mat b){
    mat ret;
    for(int i=0; i<maxn; i++){
        for(int j=0; j<maxn; j++){
            ret.m[i][j] = (ULL)0;
            for(int k=0; k<maxn; k++){
                ret.m[i][j] += a.m[i][k]*b.m[k][j];
            }
        }
    }
    return ret;
}

inline void init_unit() {
    for(int i=0; i<maxn; i++)
        unit.m[i][i] = 1;
}

mat pow_mat(mat a, long long n){
    mat ret = unit;
    while(n){
        if(n&1) ret = ret * a;
        a = a*a;
        n >>= 1;
    }
    return ret;
}

struct Aho{
    struct StateTable{
        int Next[Letter];
        int fail, flag;
    }Node[Max_Tot];
    int Size;
    queue<int> que;

    inline void init(){
        while(!que.empty()) que.pop();
        memset(Node[0].Next, 0, sizeof(Node[0].Next));
        Node[0].fail = Node[0].flag = 0;
        Size = 1;
    }

    inline void insert(char *s){
        int now = 0;
        for(int i=0; s[i]; i++){
            int idx = s[i] - ‘a‘;
            if(!Node[now].Next[idx]){
                memset(Node[Size].Next, 0, sizeof(Node[Size].Next));
                Node[Size].fail = Node[Size].flag = 0;
                Node[now].Next[idx] = Size++;
            }
            now = Node[now].Next[idx];
        }
        Node[now].flag = 1;
    }

    inline void BuildFail(){
        Node[0].fail = -1;
        for(int i=0; i<Letter; i++){
            if(Node[0].Next[i]){
                Node[Node[0].Next[i]].fail = 0;
                que.push(Node[0].Next[i]);
            }else Node[0].Next[i] = 0;///必定指向根节点
        }
        while(!que.empty()){
            int top = que.front(); que.pop();
            if(Node[Node[top].fail].flag) Node[top].flag = 1;
            for(int i=0; i<Letter; i++){
                int &v = Node[top].Next[i];
                if(v){
                    que.push(v);
                    Node[v].fail = Node[Node[top].fail].Next[i];
                }else v = Node[Node[top].fail].Next[i];
            }
        }
    }

    inline void BuildMatrix(){
        for(int i=0; i<Size; i++)
            for(int j=0; j<Size; j++)
                M.m[i][j] = 0;
        for(int i=0; i<Size; i++){
            for(int j=0; j<Letter; j++){
                if(!Node[i].flag && !Node[ Node[i].Next[j] ].flag)
                    M.m[i][Node[i].Next[j]]++;
            }
        }
        maxn = Size;
    }
}ac;

ULL GetSum(long long num){
    mat ret;
    ret.m[0][0] = 26;
    ret.m[0][1] = 1;
    ret.m[1][0] = 0;
    ret.m[1][1] = 1;
    int tmp = maxn;
    maxn = 2;
    ret = pow_mat(ret, ++num);
    maxn = tmp;
    return ret.m[0][1]-1;
}

ULL GetElimination(long long num){
    mat tmp;
    for(int i=0; i<maxn; i++)///左上角 为 原矩阵
        for(int j=0; j<maxn; j++)
            tmp.m[i][j] = M.m[i][j];

    for(int i=0; i<maxn; i++)///右上角 为 单位矩阵
        for(int j=maxn; j<(maxn<<1); j++)
            tmp.m[i][j] = (i+maxn == j);

    for(int i=maxn; i<(maxn<<1); i++)///左下角 为 零矩阵
        for(int j=0; j<maxn; j++)
            tmp.m[i][j] = 0;

    for(int i=maxn; i<(maxn<<1); i++)///右下角 为 单位矩阵
        for(int j=maxn; j<(maxn<<1); j++)
            tmp.m[i][j] = (i==j);

    int Temp = maxn;
    maxn <<= 1;///先将原本矩阵的大小放大一倍进行快速幂运算，这个和我快速幂的写法有关
    tmp = pow_mat(tmp, ++num);
    ULL ret = (ULL)0;
    maxn = Temp;///再回复成原来大小
    for(int i=maxn; i<(maxn<<1); i++)///右上角的矩阵就是幂和了
        ret += tmp.m[0][i];
        
    return (--ret);///需要 -1
}

int main(void)
{
    int n, m;

    while(~scanf("%d %d", &m, &n)){
        ac.init();
        for(int i=0; i<m; i++){
            scanf("%s", S);
            ac.insert(S);
        }
        ac.BuildFail();
        ac.BuildMatrix();
        init_unit();
        ULL Tot = GetSum((long long)n);///注意是传long long不然会爆int
        ULL Elimination = GetElimination((long long)n);
        cout<<Tot-Elimination<<endl;
    }
    return 0;
}