1003 The Dominator of Strings

Problem Description

Here you have a set of strings. A dominator is a string of the set
dominating all strings else. The string S is dominated by T if S is a
substring of T.

Input

The input contains several test cases and the first line provides the
total number of cases. For each test case, the first line contains an
integer N indicating the size of the set. Each of the following N
lines describes a string of the set in lowercase. The total length of
strings in each case has the limit of 100000. The limit is 30MB for
the input file.

Output

For each test case, output a dominator if exist, or No if not.

Sample Input

310youbetterworsericherpoorersicknesshealthdeathfaithfulnessyoubemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness5abccdeabcdeabcdebcde3aaaaaaaaabaaaac

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulnessabcdeNo

我的代码超时！！至今不知道为什么……
现在知道了>-<,加上这句就好了cin.sync_with_stdio(false);
或者ios::sync_with_stdio(false);
当然这题还可以用ac自动机，KMP算法（我都不会| |）

#include<iostream>#include<string.h>using namespace std;string str[100001];int main(){    int t;    cin>>t;    while(t--)    {        int n,mx=0,flag=1;        cin>>n;        for(int i=0;i<n;i++)        {            cin>>str[i];            if(strlen(str[mx].c_str())<strlen(str[i].c_str()))                mx=i;        }        for(int i=0;i<n;i++)        {            if(str[mx].find(str[i])==string::npos)            {                cout<<"No"<<endl;                flag=0;                break;            }        }        if(flag)            cout<<str[mx]<<endl;    }    return 0;}

修改后：

#include<iostream>#include<string.h>using namespace std;string str[100001];int main(){    cin.sync_with_stdio(false);//修改处    int t;    cin>>t;    while(t--)    {        int n,mx=0,flag=1;        cin>>n;        for(int i=0;i<n;i++)        {            cin>>str[i];            if(strlen(str[mx].c_str())<strlen(str[i].c_str()))                mx=i;        }        for(int i=0;i<n;i++)        {            if(str[mx].find(str[i])==string::npos)            {                cout<<"No"<<endl;                flag=0;                break;            }        }        if(flag)            cout<<str[mx]<<endl;    }    return 0;}