PAT-A1151 LCA in a Binary Tree (30)

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

Given any two nodes in a binary tree, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y. where X is A and Y is the other node. If U or V is not found in the binary tree, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found..

Sample Input:

6 8
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

原题..考试的时候就觉得这题做过...A1143的BST换成BT了,之前用建树的笨办法做过一次,考试直接用了

#include <cstdio>
#include <set>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN=1e4+10;

int M,N,cnt=0,ROOT=-1;
int InOrder[MAXN],PreOrder[MAXN];
set<int> st;
map<int,int> mp,pos;
struct node{
    int data,left,right,parent;
}BIT[MAXN];
int NewNode(int key,int parent){
    BIT[cnt].data=key;
    BIT[cnt].left=BIT[cnt].right=-1;
    BIT[cnt].parent=parent;
    pos[key]=cnt;
    return cnt++;
}
void CreateBIT(int &root,int pL,int pR,int iL,int iR,int parent){
    if(pL>pR) return;
    if(root==-1) root=NewNode(PreOrder[pL],parent);
    int k=mp[PreOrder[pL]];
    int nL=k-iL,nR=iR-k;
    CreateBIT(BIT[root].left,pL+1,pL+nL,iL,k-1,root);
    CreateBIT(BIT[root].right,pR-nR+1,pR,k+1,iR,root);
}
int LCA(int pu,int pv){
    vector<int> pathu,pathv;
    while(BIT[pu].parent!=-1){
        pathu.push_back(BIT[pu].data);
        pu=BIT[pu].parent;
    }
    pathu.push_back(BIT[pu].data);
    while(BIT[pv].parent!=-1){
        pathv.push_back(BIT[pv].data);
        pv=BIT[pv].parent;
    }
    pathv.push_back(BIT[pv].data);
    int A,i=pathu.size()-1,j=pathv.size()-1;
    while(i>=0&&j>=0&&pathu[i]==pathv[j]){
        A=pathu[i];
        --i;--j;
    }
    return A;
}
int main()
{
    int u,v;
    scanf("%d%d",&M,&N);
    for(int i=0;i<N;++i){
        scanf("%d",InOrder+i);
        st.insert(InOrder[i]);
        mp[InOrder[i]]=i;
    }
    for(int i=0;i<N;++i)
        scanf("%d",PreOrder+i);
    CreateBIT(ROOT,0,N-1,0,N-1,-1);
    for(int i=0;i<M;++i){
        scanf("%d%d",&u,&v);
        int fu=st.count(u),fv=st.count(v);
        if(fu==0||fv==0){
            if(fu==0&&fv==0)
                printf("ERROR: %d and %d are not found.\n",u,v);
            else if(fu==0)
                printf("ERROR: %d is not found.\n",u);
            else
                printf("ERROR: %d is not found.\n",v);
            continue;
        }
        int A=LCA(pos[u],pos[v]);
        if(A!=u&&A!=v)
            printf("LCA of %d and %d is %d.\n",u,v,A);
        else if(A==u)
            printf("%d is an ancestor of %d.\n",u,v);
        else
            printf("%d is an ancestor of %d.\n",v,u);
    }
    return 0;
}