2020ICPC新疆省赛 - A Chino With String

传送门：2020ICPC新疆省赛 - A

思路分析

不难写出一个$dp$式子$:dp[i][j] = max(dp[i-1][pre]+val[j])$，其中$pre$表示上一个可能的状态通过一个字符转移到$j$
这个式子很像另一个矩阵的经典运用，一个有向图，告诉起点和终点，求恰好$k$步走到终点的方案，所以可以用矩阵快速幂来优化

AC代码

// Problem: Chino With String
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/16792/A
// Memory Limit: 524288 MB
// Time Limit: 2000 ms

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (int)(x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;
#define int long long
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;

const int INF = 1e18;
const double PI = acos(-1);
const int N = 1e6+10;


struct AC_auto {
    struct T {
        int son[26],fail,word,last;
        void clear() {
            mem(son,0);
            word = fail = last = 0;
        }
    } tree[N];

    int cnt = 1;
    int val[N] ;

    int newnode() {
        ++cnt;
        tree[cnt].clear();
        return cnt;
    }

    void init() {
        cnt = 1;
        mem(val,0);
        tree[1].clear();
        tree[1].word = 1;
    }

    void ins(string s,int w) {
        int p = 1;
        for(auto i:s) {
            int c = i-'a';
            if(!tree[p].son[c]) tree[p].son[c] = newnode();
            p = tree[p].son[c];
        }
        tree[p].word = 1;
        val[p] += w;
    }

    void getfail() {
        for(int i=0; i<26; i++) tree[0].son[i] = 1;
        queue<int>q;
        q.push(1);
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            int ufail = tree[u].fail;
            tree[u].last = tree[ufail].word>0?ufail:tree[ufail].last;
            val[u]+=val[tree[u].last];
            for(int i=0; i<26; i++) {
                int v = tree[u].son[i];
                if(v) {
                    tree[v].fail = tree[ufail].son[i];
                    q.push(v);
                } else tree[u].son[i] = tree[ufail].son[i];
            }
        }

    }
} AC;

struct Mat{
    int n;
    vector<vector<int>>a;
    Mat(){}
    Mat(int nn){
        n = nn;
        a.resize(n+2);
        for(int i=1;i<=n;i++) a[i].resize(n+2);
        for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) a[i][j] = -INF;
    }
    
    Mat operator * (const Mat &B) const {
        Mat res(n);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                res.a[i][j] = -INF;
                for(int k=1;k<=n;k++){
                    res.a[i][j] = max(res.a[i][j],a[i][k]+B.a[k][j]);
                }
            }
        }
        return res;
    }
    
    Mat pow(int x){
        Mat res(n) , bas(n);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==j) res.a[i][j] = 0;
                else res.a[i][j] = -INF;
            }
        }
        bas.a = a;
        while(x){
            if(x&1) res = res*bas;
            x>>=1;
            bas = bas*bas;
        }
        a = res.a;
        return res;
    }
};



signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    cin>>n>>m;
    AC.init();
    for(int i=1;i<=m;i++){
        string x;
        int w;
        cin>>x>>w;
        AC.ins(x,w);
    }
    AC.getfail();
    int len = AC.cnt+1;
    Mat dp(len+1);
    for(int i=1;i<=len;i++){
        for(int j=0;j<26;j++){
            dp.a[i][AC.tree[i].son[j]] = AC.val[AC.tree[i].son[j]];
        }
    }
    dp.pow(n);
    int ans = -INF;
    for(int i=1;i<=len;i++) ans = max(ans,dp.a[1][i]);
    cout<<ans<<endl;





    return 0;
}