hdoj 6208 The Dominator of Strings

题目链接：The Dominator of Strings

题目大意：有一些字符串，问存不存在一个字符串使得其他所有字符都是这个串的子串

题目思路：首先我们知道这个串一定是最长得串，有相同最长长度的串判一下，然后会可以AC自动机直接扔，也可以用string的find函数找，卡的不是很严

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 2e5+10;int T,n,len,maxlen;string str[maxn],mo;int main(){    ios::sync_with_stdio(false);    cin>>T;    while(T--){        cin>>n;        maxlen = -1;len;        bool flag = false;        for(int i = 1;i <= n;i++){            cin>>str[i];            len = str[i].size();            if(maxlen < len) mo = str[i],maxlen = len;        }        for(int i = 1;i <= n;i++){            len = str[i].size();            if(maxlen == len&&mo != str[i]) {flag = true;break;}        }        int i;        for(i = 1;i <= n;i++){            if(mo.find(str[i]) == -1) {flag = true;break;}        }        if(!flag) cout<<mo<<endl;        else cout<<"No"<<endl;    }    return 0;}