[BZOJ4861][Beijing2017]魔法咒语 AC自动机+动态规划+矩阵快速幂

来源：互联网发布：算法分治法编辑：程序博客网时间：2024/06/05 09:18
对忌讳词语构建AC自动机
Fi,j表示长度为i，匹配到AC自动机第j位的合法串方案数
当状态数少的时候用矩阵转移
#include <bits/stdc++.h>#define N 5050const int mod = 1e9+7;using namespace std;typedef long long LL;int len[105],cnt,n,m,l,ans;char s[55][105];inline void inc(int &x,int y) {x=(x+y)%mod;}struct Trie{    int ch[N][27],leef[N],fail[N];    char s[N];    void add() {        scanf("%s",s+1); int n = strlen(s+1);        int p = 0;        for (int i=1;i<=n;i++) {            int c = s[i] - 'a';            if (!ch[p][c]) ch[p][c] = ++cnt;            p = ch[p][c];        }        leef[p] = 1;    }    void set() {        queue<int> q;        for (int i=0;i<26;i++) if (ch[0][i])            fail[ ch[0][i] ] = 0, q.push(ch[0][i]);        while (!q.empty()) {            int u=q.front(); q.pop();            for (int i=0;i<26;i++) if (ch[u][i]) {                int v = ch[u][i], t = fail[u];                while (t && !ch[t][i]) t = fail[t];                fail[v] = ch[t][i];                leef[v] = leef[v] || leef[ fail[v] ];                q.push(v);            }        }    }    int g(int u,char *s) {        int len = strlen(s+1);        for (int i=1;i<=len;i++) {            int c = s[i]-'a';            while (u && !ch[u][c]) u = fail[u];            if (ch[u][c]) u = ch[u][c];            if (leef[u]) return -1;        }        return u;    }}T;namespace planA{    int F[105][N];    void solve() {        F[0][0] = 1;        for (int _=0;_<l;_++)            for (int i=0;i<=cnt;i++) if (F[_][i])                for (int j=1;j<=n;j++) if (_+len[j]<=l) {                    int ni = T.g(i,s[j]);                    if (ni == -1) continue;                    inc(F[_+len[j]][ni], F[_][i]);                }        for (int _=0;_<=cnt;_++) inc(ans, F[l][_]);        cout << ans << endl;    }}namespace planB{    struct Matrix{int a,d[205][205];}A,B,id;    Matrix operator*(Matrix p1, Matrix p2) {        Matrix ret = id;ret.a = p1.a;        for (int i=0;i<=p1.a;i++)            for (int j=0;j<=p1.a;j++)                for (int k=0;k<=p1.a;k++)                    inc(ret.d[i][j], 1LL*p1.d[i][k] * p2.d[k][j] % mod);        return ret;    }    Matrix qp(Matrix A, int b) {        Matrix ret = id;        ret.a = A.a;        for (int i=0;i<=A.a;i++) ret.d[i][i] = 1;        while (b) {            if (b&1) ret = ret * A;            b >>= 1, A = A * A;        }        return ret;    }    void solve() {        int tp = cnt+1;        A.a = B.a = 2*tp-1;        A.d[0][0+tp] = 1;        for (int i=0;i<=cnt;i++) B.d[i+tp][i] = 1;        for (int i=0;i<=cnt;i++) {            for (int j=1;j<=n;j++) {                int ni = T.g(i,s[j]);                if (ni == -1) continue;                if (len[j] == 1)                    B.d[i+tp][ni+tp]++;                else                    B.d[i][ni+tp]++;            }        }        A = A * B;        B = qp(B, l);        A = A * B;        for (int i=0;i<=cnt;i++) inc(ans, A.d[0][i]);        cout << ans << endl;    }}int main() {    freopen("sorcery.in","r",stdin);    freopen("sorcery.out","w",stdout);    scanf("%d%d%d",&n,&m,&l);    for (int i=1;i<=n;i++) scanf("%s",s[i]+1), len[i] = strlen(s[i]+1);    for (int i=1;i<=m;i++) T.add();    T.set();    if (l <= 100) planA::solve(); else planB::solve();    return 0;}
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