#1490 : Tree Restoration
微软 2017春招真题
题目
There is a tree of N nodes which are numbered from 1 to N. Unfortunately, its edges are missing so we don't know how the nodes are connected. Instead we know:
- Which nodes are leaves
- The distance (number of edges) between any pair of leaves
- The depth of every node (the root's depth is 1)
- For the nodes on the same level, their order from left to right
Can you restore the tree's edges with these information? Note if node u is on the left of node v, u's parent should not be on the right of v's parent.
输入
The first line contains three integers N, M and K. N is the number of nodes. M is the depth of the tree. K is the number of leaves.
The second line contains M integers A1, A2, ... AM. Ai represents the number of nodes of depth i.
Then M lines follow. The ith of the M lines contains Ai numbers which are the nodes of depth i from left to right.
The (M+3)-th line contains K numbers L1, L2, ... LK, indicating the leaves.
Then a K × K matrix D follows. Dij represents the distance between Li and Lj.1 ≤ N ≤ 100
输出
For every node from 1 to N
output its parent. Output 0 for the root's parent.
样例输入
8 3 5
1 3 4
1
2 3 4
5 6 7 8
3 5 6 7 8
0 3 3 3 3
3 0 2 4 4
3 2 0 4 4
3 4 4 0 2
3 4 4 2 0
样例输出
0 1 1 1 2 2 4 4
分析
首先，给出的测试用例太过简单，先构建一个复杂的测试用例：
从下往上，从左往右遍历：
第六层（最下层）：
第一个节点26，其父节点一定是第五层第一个非叶子节点20。同时，在距离矩阵中增加节点20，与其他节点距离为节点26与其他节点距离－1
第二个节点27，若和26的距离为2，则其父节点也是20（×），否则，其父节点是第五层第二个非叶子节点22（√）同时，在距离矩阵中增加节点22，与其他节点距离为节点27与其他节点距离－1
第三个节点28，若和27的距离为2，则其父节点也是22（√），否则，其父节点是第五层第三个非叶子节点24（×）
......
第六层遍历结束，明确了第六层所有节点的父节点，也将第五层所有非叶子节点加入距离矩阵
第五层（倒数第二层）：
第一个节点19，其父节点一定是第四层第一个非叶子节点13。同时，在距离矩阵中增加节点13，与其他节点距离为节点19与其他节点距离－1
第二个节点20，若和19的距离为2，则其父节点也是13（√），否则，其父节点是第四层第二个非叶子节点14（×）
......
第五层遍历结束，明确了第五层所有节点的父节点，也将第五层所有非叶子节点加入距离矩阵
......
（一直遍历到第三层）
第二层：
所有节点的父节点都是根节点（第一层的唯一节点）
第一层：
根节点的父节点为0
解答
解法1：结果是Wrong Answer，然而并不知道为什么，谁能帮我看看错在哪了哇~
import java.util.Scanner;
import java.util.Map;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.concurrent.ConcurrentHashMap;
public class Main_2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();//节点数8
int m = sc.nextInt();//树深度3
int k = sc.nextInt();//叶子树5
int[] a = new int[m];//深度为i的节点数
int[][] arr = new int[m][];//每行的节点data
Node[][] nodes = new Node[m][];//每行的节点
Map<Integer, Node> nodesMap = new LinkedHashMap<>();//存节点
Map<Node, Map<Node,Integer>> distanceMap = new ConcurrentHashMap<>();//存节点间距离（叶子+非叶子）
for (int i = 0; i < m; i++) {
a[i] = sc.nextInt();
arr[i] = new int[a[i]];
nodes[i] = new Node[a[i]];
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < a[i]; j++){
arr[i][j] = sc.nextInt();
nodes[i][j] = new Node();
nodes[i][j].data = arr[i][j];
nodes[i][j].level = i + 1;
nodesMap.put(arr[i][j],nodes[i][j]);
if(i == 0){
nodes[i][j].parent = null;
}
else if(i == 1){
nodes[i][j].parent = nodes[0][0];
}
}
}
int[] l = new int[n];//叶子 3 5 6 7 8
for(int i = 0; i < k; i++){
l[i] = sc.nextInt();
nodesMap.get(l[i]).isLeaf = true;
}
int[][] d = new int[n][n];//叶子节点间距离
for (int i = 0; i < k; i++) {
Map<Node, Integer> tempMap = new ConcurrentHashMap<>();
for (int j = 0; j < k; j++){
d[i][j] = sc.nextInt();
if (l[i] < l[j]){
tempMap.put(nodesMap.get(l[j]),d[i][j]);
distanceMap.put(nodesMap.get(l[i]),tempMap);
}
}
}
//从下往上，从左往右遍历
for(int i = m - 1; i >= 2; i--){
int ii = i - 1;//上一层
int jj = 0;//上一层第0个节点
Node lastNode = null;
for(int j = 0; j < a[i]; j++){
if (lastNode != null){//不是本层第一个节点，计算该节点与本层上一个节点的距离
int distance = (distanceMap.get(lastNode) != null && distanceMap.get(lastNode).get(nodes[i][j]) != null) ? distanceMap.get(lastNode).get(nodes[i][j]) : distanceMap.get(nodes[i][j]).get(lastNode);
if (distance == 2){
nodes[i][j].parent = lastNode.parent;
continue;
}
}
while (nodes[ii][jj].isLeaf){
jj++;
}
nodes[i][j].parent = nodes[ii][jj];
//在距离矩阵中增加其父节点
Iterator it1 = distanceMap.entrySet().iterator();
while (it1.hasNext()) {
Map.Entry outerEntry = (Map.Entry) it1.next();
Node outerKey = (Node) outerEntry.getKey();
Map<Node,Integer> outerValue = (Map<Node,Integer>) outerEntry.getValue();
if(outerKey == nodes[i][j]){
Map<Node, Integer> tempMap = new ConcurrentHashMap<>();
Iterator it2 = outerValue.entrySet().iterator();
while (it2.hasNext()) {
Map.Entry entry = (Map.Entry) it2.next();
Node innerKey = (Node) entry.getKey();
Integer innerValue = (Integer) entry.getValue();
tempMap.put(innerKey,innerValue - 1);
}
distanceMap.put(nodes[i][j].parent,tempMap);
}
else{
Iterator it2 = outerValue.entrySet().iterator();
while (it2.hasNext()) {
Map.Entry entry = (Map.Entry) it2.next();
Node innerKey = (Node) entry.getKey();
Integer innerValue = (Integer) entry.getValue();
if(innerKey == nodes[i][j]){
Map<Node, Integer> tempMap = distanceMap.get(outerKey);
tempMap.put(nodes[i][j].parent,innerValue - 1);
distanceMap.put(outerKey,tempMap);
}
}
}
}
lastNode = nodes[i][j];
jj++;
}
}
for (Map.Entry<Integer, Node> entry : nodesMap.entrySet()) {
Node thisNode = entry.getValue();
int result = (thisNode.parent != null) ? thisNode.parent.data : 0;
System.out.print(result+" ");
}
}
}
class Node{
public Node parent;
public int data;
public int level;
public boolean isLeaf;
}
附两个测试用例：
9 4 6
1 2 3 4
9
1 10
2 3 4
5 6 7 8
3 5 6 7 8 10
0 3 3 3 3 3
3 0 2 4 4 4
3 2 0 4 4 4
3 4 4 0 2 4
3 4 4 2 0 4
3 4 4 4 4 0
31 6 13 1 4 6 7 7 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3 8 12 15 19 21 23 26 27 28 29 30 31 0 3 4 4 5 5 5 6 6 6 6 6 6 3 0 3 5 4 4 6 5 7 7 7 7 7 4 3 0 6 5 5 7 6 8 8 8 8 8 4 5 6 0 7 7 7 8 4 4 8 8 8 5 4 5 7 0 4 8 3 9 9 9 9 9 5 4 5 7 4 0 8 5 9 9 9 9 9 5 6 7 7 8 8 0 9 9 9 7 7 7 6 5 6 8 3 5 9 0 10 10 10 10 10 6 7 8 4 9 9 9 10 0 2 10 10 10 6 7 8 4 9 9 9 10 2 0 10 10 10 6 7 8 8 9 9 7 10 10 10 0 2 4 6 7 8 8 9 9 7 10 10 10 2 0 4 6 7 8 8 9 9 7 10 10 10 4 4 0
