Codeforces Round #434 D

Polycarp's phone book

题意：给n个号码，求出每个号码最短的能唯一定位该号码的字符串

思路：暴力 把每个号码的子串放进map里，然后从长度短的开始暴力每一个子串出现的次数，出现一次的就是能定位的，同一个号码里的相同子串只记录一次，如0000 ，子串000出现2次，但是只记录一次（当然也可以字典树或者后缀数组写拉，但是暴力是坠吼滴

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

string s[70002];
map<string,int> M,m[70002];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n; cin>>n;
    for(int i=1; i<=n; ++i){
        cin>>s[i];
    }
    for(int i=1; i<=n; ++i){
        for(int j=0; j<9; ++j){
            for(int k=1; k+j<=9; ++k){
                string ss=s[i].substr(j, k);
                if(!m[i][ss]){
                    m[i][ss]++;
                    M[ss]++;
                }
            }
        }
    }
    for(int i=1; i<=n; ++i){
        int flag=0;
        for(int k=1; k<=9; ++k){
            for(int j=0; j+k<=9; ++j){
                string ss=s[i].substr(j, k);
                if(M[ss]==1){
                    cout<<ss<<endl;
                    flag=1; break;
                }
            }
            if(flag) break;
        }
    }
    return 0;
}