原题来自Careercup: http://www.careercup.com/question?id=5840928073842688

We have a list of N nodes with each node pointing to one of the N nodes.

It could even be pointing to itself. We call a node ‘good’,

if it satisfies one of the following properties:

* It is the tail node (marked as node 1)

* It is pointing to the tail node (node 1)

* It is pointing to a good node

You can change the pointers of some nodes in order to make them all ‘good’.

You are given the description of the nodes.

You have to find out what is minimum number of nodes that you have to change in order

to make all the nodes good.

Input:

The first line of input contains an integer number N which is the number of nodes.

The next N lines contains N numbers,

all between 1 and N.

The first number, is the number of the node pointed to by Node 1;

the second number is the number of the node pointed to by Node 2;

the third number is the number of the node pointed to by Node 3 and so on.

N is no larger than 1000.

Output:

Print a single integer which is the answer to the problem

Sample Input 1:

5

1

2

3

4

5

Sample output 1:

4

Explanation:

We have to change pointers for four nodes (node #2 to node #5) to point to node #1.

Thus 4 changes are required

Sample input 2:

5

5

5

5

5

5

Sample output 2:

1

Explanation:

We have to just change node #5 to point to node #1 (tail node) which will make node #5 good.

Since all the other nodes point to a good node (node #5), every node becomes a good node.

if it satisfies one of the following properties:

* It is the tail node (marked as node 1)

* It is pointing to the tail node (node 1)

* It is pointing to a good node

You can change the pointers of some nodes in order to make them all ‘good’.

You are given the description of the nodes.

You have to find out what is minimum number of nodes that you have to change in order

to make all the nodes good.

Input:

The first line of input contains an integer number N which is the number of nodes.

The next N lines contains N numbers,

all between 1 and N.

The first number, is the number of the node pointed to by Node 1;

the second number is the number of the node pointed to by Node 2;

the third number is the number of the node pointed to by Node 3 and so on.

N is no larger than 1000.

Output:

Print a single integer which is the answer to the problem

Sample Input 1:

5

1

2

3

4

5

Sample output 1:

4

Explanation:

We have to change pointers for four nodes (node #2 to node #5) to point to node #1.

Thus 4 changes are required

Sample input 2:

5

5

5

5

5

5

Sample output 2:

1

Explanation:

We have to just change node #5 to point to node #1 (tail node) which will make node #5 good.

Since all the other nodes point to a good node (node #5), every node becomes a good node.

另外还需考虑4 4 3 2 1为输入的情况，输出应该是1.

这题用union-find(并查集，或者叫disjoint-set)来做。

关于并查集，参考http://blog.csdn.net/dm_vincent/article/details/7655764

**Code (Java):**

import java.util.*; public class Solution { public static void main(String[] args) throws Exception { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); int[] nodes = new int[n]; for (int i = 0; i < n; i++) { nodes[i] = scanner.nextInt() - 1; } int result = minimumConnects(nodes); System.out.println(result); scanner.close(); } public static int minimumConnects(int[] nodes) { if (nodes == null || nodes.length == 0) { return 0; } int n = nodes.length; int[] parents = new int[n]; for (int i = 0; i < n; i++) { parents[i] = i; } // Union for (int i = 0; i < n; i++) { union(parents, i, nodes[i]); } // Calculate the minimum connections needed int count = 0; for (int i = 1; i < n; i++) { if (parents[i] == i) { count++; } } return count; } private static void union(int[] parents, int p, int q) { int pRoot = find(parents, p); int qRoot = find(parents, q); parents[pRoot] = qRoot; } private static int find(int[] parents, int p) { if (p == 0) { return 0; } while (p != parents[p]) { p = parents[p]; } return p; } }

不太理解 为什么 44 231 的时候输出是1，觉得是2， 能解释下嘛？谢谢啦。

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