UVa 1399 Puzzle

方法：AC自动机

先把禁止的string插入ac自动机中，然后再这个自动机上求最长的合法路径。如果出现环或者最长路径长度为0，则输出“No"；否则输出最长路径。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

const int maxnode = 4e5+5, sigma_size = 26;
int n, K;
struct AC
{
    int sz, g[maxnode][sigma_size];
    int fail[maxnode], last[maxnode];
    void init()
    {
        sz = 1; Reset(g[0], 0);
    }
    int idx(char c)
    {
        return c-'A';
    }
    int newnode()
    {
        last[sz] = 0; Reset(g[sz], 0);
        return sz++;
    }
    void insert(const string &str)
    {
        int n = str.length(), cur = 0;
        rep(i, n)
        {
            int c = idx(str[i]);
            if (!g[cur][c]) g[cur][c] = newnode();
            cur = g[cur][c];
        }
        last[cur] = 1;
    }
    void get_fail()
    {
        queue<int> q;
        fail[0] = 0;
        rep(i, sigma_size)
        {
            int u = g[0][i];
            if (u)
            {
                fail[u] = 0;
                q.push(u);
            }
        }
        while (!q.empty())
        {
            int cur = q.front();  q.pop();
            rep(i, K)
            {
                int u = g[cur][i];
                if (!u)
                {
                    g[cur][i] = g[fail[cur]][i]; continue;
                }
                q.push(u);  int nxt = fail[cur];
                while (nxt && !g[nxt][i]) nxt = fail[nxt];
                fail[u] = g[nxt][i];
                last[u] |= last[fail[u]];
            }
        }
    }
} solver;
bitset<maxnode> vis;
int dp[maxnode], jump[maxnode];

int dfs(int cur)
{
    if (vis[cur]) return -1;
    int &ans = dp[cur];
    if (ans == -1)
    {
        vis[cur] = true;
        ans = 0;
        drep(i, K)
        {
            int nxt = solver.g[cur][i];
            if (!solver.last[nxt])
            {
                int tmp = dfs(nxt);
                if (tmp == -1) return -1;
                if (tmp+1 > ans)
                {
                    ans = tmp+1;
                    jump[cur] = i;
                }
            }
        }
    }
    vis[cur] = 0;
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> K >> n;
        solver.init();
        Reset(dp, -1);
        Reset(jump, -1);
        vis.reset();
        string line;
        rep(i, n)
        {
            cin >> line;
            solver.insert(line);
        }
        solver.get_fail();
        int ans = dfs(0);
        if (ans == -1 || !ans)
        {
            cout << "No
";
        }
        else
        {
            
            int cur = 0;
            for(;;)
            {
                if (jump[cur] == -1) break;
                cout << (char)(jump[cur]+'A');
                cur = solver.g[cur][jump[cur]];
            }
            cout << newline;
        }
    }
}

/*
 
 3 2 4 AAA AB BA BB 2 4 AAA BBB ABAB BBAA 3 7 AA ABA BAC BB BC CA CC

*/