题目链接
https://pintia.cn/problem-sets/994805046380707840/problems/994805066135879680
思路
对于朋友和朋友之间的关系我们通过并查集维护即可,然后对于敌人之间的关系,我们只需要用一个二维数组或者使用map就能维护,因为只有直接敌对关系才是敌人,然后对于每一次询问我们就根据他们俩的朋友关系、敌对关系分类讨论即可
代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f
const int N = 1e2+10;
int fa[N],n,m,k;
bool enmy[N][N];
int find(int x){
return x==fa[x]?x:find(fa[x]);
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin>>n>>m>>k;
for(int i = 1;i <= n; ++i) fa[i] = i;
int u,v,w;
for(int i = 1;i <= m; ++i) {
cin>>u>>v>>w;
if(w + 1) {
u = find(u);
v = find(v);
fa[v] = u;
} else {
enmy[u][v] = enmy[v][u] = true;
}
}
for(int i = 1;i <= k; ++i) {
cin>>u>>v;
if(enmy[u][v]) {
if(find(u) == find(v)) cout<<"OK but..."<<endl;
else cout<<"No way"<<endl;
} else {
if(find(u) == find(v))
cout<<"No problem"<<endl;
else cout<<"OK"<<endl;
}
}
return 0;
}