M
1520320968
tags: Union Find
Lint还不能跑, 全部按照题意和答案document的.
```
/*
LintCode:
Given n nodes in a graph labeled from 1 to n. There is no edges in the graph at beginning.
You need to support the following method:
1. connect(a, b), add an edge to connect node a and node b.
2. query(a), Returns the number of connected component nodes which include node a.
Example
5 // n = 5
query(1) return 1
connect(1, 2)
query(1) return 2
connect(2, 4)
query(1) return 3
connect(1, 4)
query(1) return 3
*/
/**
Thoughts:
No chance to run on LintCode.
*/
public class ConnectingGraph {
// Placeholder for all the UninFind relationships
private int[] father = null;
// Stores the number of union elements connected from downstream to each individual node
private int[] size = null;
/**
Initialize one element to each individual union; size will be 1 for all unions as well.
*/
public ConnectingGraph(int n) {
father = new int[n + 1];
size = new int[n + 1];
for (int i = 1; i <= n; i++) {
father[i] = i;
size[i] = 1;
}
}
/**
Union function.
Find the root father, if not the same, union them together.
*/
public void connect(int a, int b) {
int rootA = find(a);
int rootB = find(b);
if (rootA != rootB) {
parent[a] = rootB; // doesn't mater which assigns to which one.
// Merge union of A into the union of B, so unionB should grow.
size[rootB] += size[rootA]
}
}
/**
Returns the number of connected component nodes which include node a.
*/
public boolean query(int x) {
int rootX = find(x);
return size[rootX];
}
/*
Find function: find the root parent as the head for entire union.
If found parent as itself, return it.
Otherwise, recursively look for father and assign the result eventually.
*/
private int find(int x) {
if (father[x] == x) {
return x; // x is the root parent, return itself.
}
return father[x] = find(father[x]);
}
}
```