codeforces 1360F 思维

思维        

题意：q个测试样例。n个字符串，找到一个字符串满足该字符串与所有的字符串最多只差一个字符。

思路：n，m范围很小，遍历n个字符串，找出他们的衍生字符串，即每位都变化26次。出现了n次的就是答案。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int main()
{
    //freopen("input.txt", "r", stdin);
    int q;
    scanf("%d",&q);
    //set<pair<string,int> >st;
    
    
    while(q--)
    {
        int n,m;
        cin>>n>>m;
        map<string,int>mp;
        rep(i,1,n)
        {
            string s;
            cin>>s;
            set<string>st;
            rep(j,0,m-1)
            {
                char c = s[j];
                rep(k,97,122)
                {
                    s[j] = char(k);
                    st.insert(s);
                }
                s[j] = c;
            }
            for(auto i : st){
                mp[i]++;
            }
        }
        string ans;
        int flag = 0;
        for(auto i : mp){
            int t = i.se;
            if(t == n){
                flag = 1;
                ans = i.fi;
            }

/*            if(st.count(i) == n){
                ans = i;
                flag = 1;
                break;
            }*/
        }
        if(flag == 1){
            cout<<ans<<endl;
        }
        else cout<<-1<<endl;
    }
}