HDU 6208 The Dominator of Strings(字符串）

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 887 Accepted Submission(s): 273

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
3
10
you
better
worse
richer
poorer
sickness
health
death
faithfulness
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
5
abc
cde
abcde
abcde
bcde
3
aaaaa
aaaab
aaaac

Sample Output
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
abcde
No
題意：在n個字符串中找到一個串。其他串都是它的子串。沒有輸出No。
題解：string.find暴力判斷。。
代碼：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<algorithm>
#include<map>
#include<time.h>
#include<math.h>
//#define pb push_back
//#define mp make_pair
#define INF 0x3f3f3f3f
using namespace std;
typedef long long int ll  ;
typedef pair<int,int>pp;
const int N=1e5+100;
const ll mod=1e9+7;
int read()
{
    int x=0;
    char ch = getchar();
    while('0'>ch||ch>'9')ch=getchar();
    while('0'<=ch&&ch<='9')
    {
        x=(x<<3)+(x<<1)+ch-'0';
        ch=getchar();
    }
    return x;
}
/***********************************************************/
string str[N];
int t,n;
int main()
{
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--)
    {
        bool flag=true;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>str[i];
        }
        for(int i=1;i<=n;i++)
        {
            int ans=0;
            for(int j=1;j<=n;j++)
            {
                if(str[i].find(str[j])!=-1) ans++;
                    else break;
            }
            if(ans==n)
            {
                cout<<str[i]<<endl;
                flag=false;
                break;
            }
        }
        if(flag) cout<<"No"<<endl;
    }
    return 0;
}