C - Not Afraid

CodeForces - 787B

Since the giant heads have appeared in the sky all humanity is in danger, so allRicks and Mortys from all parallel universes are gathering in groups to find a solution to get rid of them.
There are n parallel universes participating in this event (n Ricks and n Mortys). I. e. each of n universes has one Rick and one Morty. They're gathering in m groups. Each person can be in many groups and a group can contain an arbitrary number of members.
Ricks and Mortys have registered online in these groups. So, a person can have joined a group more than once (developer of this website hadn't considered this possibility).

Input
The first line of input contains two integers n and m (1?≤?n,?m?≤?104
) — number of universes and number of groups respectively.
The next m lines contain the information about the groups. i-th of them first contains an integer k (number of times someone joined i-th group, k?>?0) followed by k integers vi,?1,?vi,?2,?...,?vi,?k. If vi,?j is negative, it means that Rick from universe number ?-?vi,?j has joined this group and otherwise it means that Morty from universe number vi,?jhas joined it.Sum of k for all groups does not exceed 104.

Output
In a single line print the answer to Summer's question. Print "YES" if she should cancel the event and "NO" otherwise.

Example
Input
4 21 -34 -2 3 2 -3

Output
YES

Input
5 25 3 -2 1 -1 53 -5 2 5

Output
NO

Input
7 23 -1 6 77 -5 4 2 4 7 -3 4

Output
YES

Note
In the first sample testcase, 1st group only contains the Rick from universe number 3, so in case he's a traitor, then all members of this group are traitors and so Summer should cancel the event.

題意：如果每一組中至少存在一對相反數(shù)則輸出no，否則輸出yes

解法：用set來存儲，對于一個數(shù)n如果-n在set中，則這一組合格藻烤，所有組都合格則輸出No

代碼：

#include<iostream>
#include<set>
using namespace std;
int main()
{
    int n,m,k;
    cin>>n>>m;
    set<int> s;
    int ans=1;
    for(int i=0;i<m;i++){
        s.clear();
        cin>>k;
        int v;
        int flag=0;
        for(int j=0;j<k;j++){
            cin>>v;
            if(s.count(-v))
                flag=1;
            s.insert(v);
        }
        ans=ans&flag;
    }
    ans?cout<<"NO"<<endl:cout<<"YES"<<endl;
    return 0;
}