POJ

题目连接：https://vjudge.net/problem/POJ-2778

题意：给m个病毒串，问不包含病毒串且长度n的DNA片段有几个。

题解：我看了好多个题解讲的都不是很清楚，终于找到一个看明白的了。

https://blog.csdn.net/morgan_xww/article/details/7834801

（1）用病毒串构造一个AC自动机，病毒串末尾的结点要标记，如果当前结点fail指向的结点被标记了，那么当前节点也要被标记。

（2）用AC自动机构造矩阵a，表示到达某个状态的合法转移次数。

（3）求矩阵的n次幂，第0行的和即为结果；

#include<iostream>
#include<algorithm>
#include<queue>
#include<string.h>
#define ll long long
using namespace std;

const ll mod=1e5;
const ll maxn = 111;
const ll MX = 111;
ll ans=0;
ll d['Z'+1];

struct tire{
    ll nxt[maxn][4],fail[maxn],end[maxn],tot,root,vis[maxn];
    ll newNode(){
        for(ll i=0;i<4;i++) nxt[tot][i] = -1;
        end[tot++] = 0;
        return tot-1;
    }
    void Init(){
        tot = 0;
        root = newNode();
        ans=0;
    }
    void Insert(char *buf){
        ll len = strlen(buf),i,u = root;
        for(i=0;i<len;i++){
            ll x = d[buf[i]];
            if(nxt[u][x]==-1) nxt[u][x] = newNode();
            u = nxt[u][x];
        }
        end[u] = 1;
    }
    void build(){
        queue <ll> q;
        fail[root] = root;
        for(ll i=0;i<4;i++){
            if(nxt[root][i]==-1) nxt[root][i] = root;
            else{
                fail[nxt[root][i]] = root;
                q.push(nxt[root][i]);
            }
            end[nxt[root][i]]|=end[fail[nxt[root][i]]];
        }
        while(!q.empty()){
            ll now = q.front();
            q.pop();
            for(ll i=0;i<4;i++){
                if(nxt[now][i]==-1) nxt[now][i] = nxt[fail[now]][i];
                else{
                    fail[nxt[now][i]] = nxt[fail[now]][i];
                    q.push(nxt[now][i]);
                }
                end[nxt[now][i]]|=end[fail[nxt[now][i]]];
            }
        }
    }
}ac;

struct Mat{
    ll mat[maxn][maxn];
};

Mat operator * (Mat a, Mat b) {
    Mat c;
    ll i, j, k;
    memset(c.mat, 0, sizeof(c.mat));
    for (i = 0; i <= ac.tot; i++)
        for (j = 0; j <= ac.tot; j++)
            for (k = 0; k <= ac.tot; k++) {
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                c.mat[i][j] %= mod;
            }
    return c;
}

Mat operator ^ (Mat a, ll b) {
    Mat c;
    ll i, j;
    for (i = 0; i <= ac.tot; i++)
        for (j = 0; j <= ac.tot; j++)
            c.mat[i][j] = (i == j);
    while(b != 0) {
        if (b & 1 != 0) c = c * a;
            a = a * a;
            b /= 2;
    }
    return c;
}

char s1[MX];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t,n,m;
    cin>>n>>m;
    d['A']=0,d['C']=1,d['T']=2,d['G']=3;
    ac.Init();
    for(ll i=0;i<n;i++) cin>>s1,ac.Insert(s1);
    ac.build();
    Mat a,b;
    for(ll i=0;i<=ac.tot;i++){
        if(ac.end[i]) continue;
        for(ll j=0;j<4;j++){
            if(ac.end[ac.nxt[i][j]]) continue;
            a.mat[i][ac.nxt[i][j]]++;
        }
    }
    b=a^m;
    for(ll i=0;i<=ac.tot;i++){
        ans+=b.mat[0][i];
        ans%=mod;
    }
    cout<<ans<<endl;
    return 0;
}