M

1520323016

tags: Union Find

还是UnionFind的变形, 这次是算有剩下多少个union. 其实非常简单, 维持一个全局变量count:

一开始count=n, 因为全是散装elements; 每次union, count--.

```

/*

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(), Returns the number of connected component in the graph

Example

5 // n = 5

query() return 5

connect(1, 2)

query() return 4

connect(2, 4)

query() return 3

connect(1, 4)

query() return 3

*/

/**

Thoughts:

Not able to run on LintCode.

Count the number of unions left

*/

public class ConnectingGraph {

// Placeholder for all the UninFind relationships

private int[] father = null;

private int count;

/**

Initialize one element to each individual union.

*/

public ConnectingGraph(int n) {

father = new int[n + 1];

count = n;

for (int i = 1; i <= n; i++) {

father[i] = i;

}

}

/**

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) {

count--;

parent[a] = rootB; // doesn't mater which assigns to which one.

}

}

/** Return count of union left */

public boolean query() {

return count;

}

/*

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]);

}

}

```