HDU 6208 The Dominator of Strings（stl）

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 2439    Accepted Submission(s): 879

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
        

Source 2017 ACM/ICPC Asia Regional Qingdao Online  

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題解：

很水的一題，暴力就可以解決，直接取最長的作為模闆串，如果和他等長的串和他不相同就不可以，然後其他的串要在那個串裡面找得到。。。

比賽的時候我tm也是醉了，莫名其妙wa了好幾遍，最後也沒發現錯，後來比完賽才知道錯在了取消了cin和cout的同步以後用printf就會出錯了。。。我把printf改成cout就ac了，我%@#￥@#￥@#￥@，以後不能随便用取消同步了

代碼：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
#define ll long long
#define INF 100861111
#define M (t[k].l+t[k].r)/2
#define lson k*2
#define rson k*2+1
using namespace std;
struct node
{
    string s;
};
node p[100005];
string t;
int main()
{
    std::ios::sync_with_stdio(false);
    int test,i,j,n,num,maxx,tag;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d",&n);
        maxx=0;
        for(i=0;i<n;i++)
        {
            cin>>p[i].s;
            if(p[i].s.size()>maxx)
            {
                maxx=p[i].s.size();
                t=p[i].s;
            }
        }
        tag=1;
        for(i=0;i<n;i++)
        {
            if(p[i].s.size()==maxx)
            {
                if(p[i].s!=t)
                {
                    tag=0;
                    break;
                }
            }
            else
            {
                if(t.find(p[i].s)==string::npos)
                {
                    tag=0;
                    break;
                }
            }
        }
        if(tag)
            cout<<t<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}