【解题参考blog】http://blog.csdn.net/acm_10000h/article/details/48878645

【题意】给定K和N，表示有K种不同的字符，N个禁止串，求一个最长的串使得该串不包含任何禁止串为子串。如果存在循环或者不能构成的话，输出No！

【解题方法】实际上就是Trie图上的最长路，先对模式串建立AC自动机，之后在AC自动机上跑最长路。无限长的情况就是trie图形成了环，这个dfs判环即可！

【代码君】

//
//Created by just_sort 2016/10/13
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 55555;
const int maxm = 26;
struct Acautomata{
    int ch[maxn][maxm],val[maxn],fail[maxn],root,sz;
    int newnode(){
        val[sz] = 0;
        memset(ch[sz], 0, sizeof(ch[sz]));
        return sz++;
    }
    void init(){
        sz = 0;
        root = newnode();
    }
    void insert(char *s,int v){
        int len = strlen(s);
        int u = root;
        for(int i = 0; i < len; i++){
            int now = s[i] - 'A';
            if(!ch[u][now]){
                ch[u][now] = newnode();
            }
            u = ch[u][now];
        }
        val[u] = v;
    }
    void build(){
        queue <int> q;
        fail[0] = 0;
        for(int i = 0; i < maxm; i++){
            int u = ch[0][i];
            if(u){
                q.push(u);
                fail[u] = 0;
            }
        }
        while(q.size())
        {
            int u = q.front(); q.pop();
            for(int i = 0; i < maxm; i++){
                int v = ch[u][i];
                if(!v){
                    ch[u][i] = ch[fail[u]][i]; continue;
                }
                q.push(v);
                int j = fail[u];
                while(j && !ch[j][i]) j = fail[j];
                fail[v] = ch[j][i];
                val[v] |= val[fail[v]];
            }
        }
    }
}ac;
int n,m;
int vis1[maxn],vis2[maxn],dp[maxn],path[maxn][2];

bool dfs1(int u)
{
    vis2[u] = 1;
    for(int i = 0; i < n; i++)
    {
        int v = ac.ch[u][i];
        if(vis1[v]) return 1;
        if(!vis2[v] && !ac.val[v])
        {
            vis1[v] = 1;
            if(dfs1(v)) return true;
            vis1[v] = 0;
        }
    }
    return false;
}

int dfs2(int u)
{
    if(vis1[u]) return dp[u];
    vis1[u] = 1;
    dp[u] = 0;
    for(int i = n - 1; i >= 0; i--)
    {
        if(!ac.val[ac.ch[u][i]])
        {
            int tmp = dfs2(ac.ch[u][i]) + 1;
            if(dp[u] < tmp)
            {
                dp[u] = tmp;
                path[u][0] = ac.ch[u][i];
                path[u][1] = i;
            }
        }
    }
    return dp[u];
}

void print_path(int u)
{
    if(path[u][0] == -1) return ;
    printf("%c", 'A' + path[u][1]);
    print_path(path[u][0]);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ac.init();
        scanf("%d%d",&n,&m);
        for(int i = 0; i < m; i++){
            char op[55];
            scanf("%s",op);
            ac.insert(op,1);
        }
        ac.build();
        memset(vis1, 0, sizeof(vis1));
        memset(vis2, 0, sizeof(vis2));
        vis1[0] = 1;
        if(dfs1(0))
        {
            printf("No\n");
        }
        else{
            memset(vis1, 0, sizeof(vis1));
            memset(path, -1, sizeof(path));
            if(dfs2(0) == 0)
            {
                printf("No\n");
            }
            else
            {
                print_path(0);
                puts("");
            }
        }
    }
    return 0;
}