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还是比较简单的一道题，思路就是并查集 + map<int,map<int,int>>，其中并查集用来把所有是朋友的放到一起，map用来存储两个人是否是敌人

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
typedef long long ll;
const int N = 110;
const int mod = 998244353;
int n,m,k;
int fa[N];
int find(int x){if(fa[x] == x){return x;}return fa[x] = find(fa[x]);
}
void merge(int x,int y){int xx = find(x),yy = find(y);if(xx == yy){return;}fa[yy] = xx;
}
void solve() {cin>>n>>m>>k;for(int i = 0 ; i <= 100 ; i++){fa[i] = i;}map<int,map<int,int>> mp;	//记录两个人是不是死对头for(int i = 0 ; i < m ; i++){int x,y,op;cin>>x>>y>>op;if(op == 1){merge(x,y);}else{mp[x][y] = 1;mp[y][x] = 1;}}for(int i = 0 ; i<k ; i++){int x,y;cin>>x>>y;if(mp[x][y] == 0 && find(x) == find(y)){cout<<"No problem"<<endl;}else if(mp[x][y] == 0 && find(x) != find(y)){cout<<"OK"<<endl;}else if(mp[x][y] == 1 && find(x) == find(y)){cout<<"OK but..."<<endl;}else{cout<<"No way"<<endl;}}
}
signed main() {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);int tt = 1;//    cin >> tt;while (tt--) {solve();}return 0;
}