L2-010 row seats (25 points) and check the collection

Topic details - L2-010 Row of seats (25 branch ) (pintia.cn)

One key point is to judge whether two people have common friends , This equivalence is related to whether the graph is connected , Therefore, we use parallel search set to solve the problem

#include<iostream>
using namespace std;
const int N = 110;
int a[N][N],f[N];
int n,m,k;
bool check(int x,int y){
    for(int i = 1; i <= n; i++){
        if(a[x][i] == 1 && a[y][i] == 1){
            return true;
        }
    }
    return false;
}

int getf(int x){
    if(x == f[x]) return x;
    else return f[x] = getf(f[x]);
}

void merge(int a,int b){
    int fa = getf(a);
    int fb = getf(b);
    if(fa != fb) f[fb] = fa;
}

int main(){
    
    cin>>n>>m>>k;
    for(int i = 1; i <= n; i++) f[i] = i;
    for(int i = 1; i <= m; i++){
        int u,v,w;
        cin>>u>>v>>w;
        a[u][v] = a[v][u] = w;
        if(w == 1){
            merge(u,v);
        }
    }
    
    while(k--){
        int x,y;
        cin>>x>>y;
        if(a[x][y] == 1) puts("No problem");
        else if(a[x][y] == 0) puts("OK");
        else {
            if(getf(x) == getf(y)) puts("OK but...");
            else puts("No way");
        }
    }
    return 0;
}