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POJ 2778: DNA Sequence

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题意

求构造不含m(0 <= m <= 10)个模式串的长度为n(1 <= n <= 2000000000)的字符串的方案数。

分析

首先用m个模式串构造AC自动机，这里需要注意的是，不仅要把字符串最后一个字符所在的那一点标记为不合法状态，还需要把其他能转移到该点的节点都标记为不合法状态。然后，问题就转化为了从root出发，走了n步，这n步都避开不合法状态的方案数。设AC自动机的节点个数为L，我们可以构造一个L*L的邻接矩阵，mat[i][j]代表i到j的合法方案数。则mat矩阵的n次幂就表示走了n次的方案数。由于n很大，所以要用矩阵快速幂优化。这里要注意的是，每次循环都要将不合法的项置0，否则，有可能会多算。

代码

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#pragma comment(linker, "/STACK:102400000,102400000")    //手动扩栈

#include<set>
#include<map>
#include<ctime>        //CLOCKS_PER_SEC;clock_t t=clock();
#include<cmath>        
#include<queue>
#include<bitset>
#include<cctype>
#include<cstdio>
#include<vector>
#include<string>    //getline(cin, line);
#include<sstream>    //stringstream ss(line);(ss is a stream like cin).
#include<cstdlib>
#include<cstring> 
#include<cfloat>    //X=FLT,DBL,LDBL;X_MANT_DIG,X_DIG,X_MIN_10_EXP,X_MIN_10_EXP,X_MIN,X_MAX,X_EPSILON
#include<climits>    //INT_MAX,LLONG_MAX
#include<iostream>     //ios_base::sync_with_stdio(false);
#include<algorithm>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double PI = acos(-1.0);
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const int N=1e5+9;
const int shift=1e3+9;
const double Eps=1e-7;

const int mod =    100000;
map<char, int> m;
int n, mm;
char s[19];

struct acAuto {
    int next[109][4], fail[109], L, root;
    ll mat[109][109];
    bool end[109];
    void init() {
        L = 0;
        m['A'] = 0;
        m['T'] = 1;
        m['C'] = 2;
        m['G'] = 3;
        root = newNode();
    }
    int newNode() {
        for(int i = 0; i < 4; i++)
            next[L][i] = -1;
        end[L] = false;
        return L++;
    }
    void insert(char s[]) {
        int len = strlen(s), now = root;
        for(int i = 0; i < len; i++) {
            int j = m[s[i]];
            if(next[now][j] == -1)
                next[now][j] = newNode();
            now = next[now][j];
        }
        end[now] = true;
    }
    void build() {
        fail[root] = root;
        queue<int> que;
        for(int i = 0; i < 4; i++) {
            if(next[root][i] == -1)
                next[root][i] = root;
            else {
                fail[next[root][i]] = root;
                que.push(next[root][i]);
            }
        }
        while(que.size()) {
            int now = que.front();
            que.pop();
            for(int i = 0; i < 4; i++) {
                if(next[now][i] == -1) {
                    next[now][i] = next[fail[now]][i];
                }
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    que.push(next[now][i]);
                }
            }
        }
    }
    void buildMatrix() {
        memset(mat, 0, sizeof mat);
        for(int i = 0; i < L; i++) {
            int now = i, flag = false;
            while(now != root) {
                if(end[now]) flag = true;
                now = fail[now];
            }
            if(flag)
                end[i] = true;
        }
        for(int i = 0; i < L; i++)
            for(int j = 0; j < 4; j++)
                mat[i][next[i][j]]++;
    }
    void ksm(int n) {
        ll t[109][109];
        ll t1[109][109];
        ll t2[109][109];
        memset(t, 0, sizeof t);
        for(int i = 0; i < L; i++)
            t[i][i] = 1;
        while(n) {
            for(int j = 0; j < L; j++)
                if(end[j] == true)
                    for(int i = 0; i < L; i++)
                        mat[i][j] = t[i][j] = 0;
            if(n & 1) {
                memcpy(t1, t, sizeof t);
                memset(t, 0, sizeof t);
                for(int i = 0; i < L; i++)
                    for(int j = 0; j < L; j++)
                        for(int k = 0; k < L; k++)
                            (t[i][j] += t1[i][k] * mat[k][j]) %= mod;
            }
            n >>= 1;
            memcpy(t2, mat, sizeof mat);
            memset(mat, 0, sizeof mat);
            for(int i = 0; i < L; i++)
                for(int j = 0; j < L; j++)
                    for(int k = 0; k < L; k++)
                        (mat[i][j] += t2[i][k] * t2[k][j]) %= mod;
        }
        for(int i = 0; i < L; i++)
            for(int j = 0; j < L; j++)
                mat[i][j] = t[i][j];
    }
    void solve() {
        buildMatrix();
        ksm(n);
        ll ans = 0;
        for(int j = 0; j < L; j++)
            (ans += mat[0][j]) %= mod;
        printf("%lld\n", ans);
    }
}ac;

int main(void) {
    if(fopen("in", "r")!=NULL) {freopen("in", "r", stdin); freopen("out", "w", stdout);}
    while(~scanf("%d%d", &mm, &n)) {
        ac.init();
        for(int i = 0; i < mm; i++) {
            scanf("%s", s);
            ac.insert(s);
        }
        ac.build();
        ac.solve();
    };
    return 0;
}