1143 Lowest Common Ancestor

link

 

解法：

本题是bst，无需建树。按前序遍历的顺序检查u，v是否在当前点的左右即可。

 

int M,N;
unordered_set<int> intree;


int pre[10002];
int main(){
    cin>>M>>N;

    for(int i=1;i<=N;i++){
        cin>>pre[i];
        intree.insert(pre[i]);

    }
    while(M--){
        int u,v;
        cin>>u>>v;
        auto f1=intree.find(u);
        auto f2=intree.find(v);
        if(f1==intree.end() && f2==intree.end()){
            printf("ERROR: %d and %d are not found.\n", u,v);
        }else if(f1==intree.end()){
            printf("ERROR: %d is not found.\n", u);
        }else if(f2==intree.end()){
            printf("ERROR: %d is not found.\n", v);
        }else {
            int res=0;
            for(int i=1;i<=N;i++){
                if((u<=pre[i] && v>=pre[i])||(u>=pre[i] && v<=pre[i])){
                    res=pre[i];
                    break;
                }
            }
            if(res==u){
                printf("%d is an ancestor of %d.\n", u,v);
            }else if(res==v){
                printf("%d is an ancestor of %d.\n", v,u);
            }else {
                printf("LCA of %d and %d is %d.\n", u,v,res);
            }
        }
    }
    return 0;
}