Max-Spacing k-clustering

这里主要用到的知识点是Union-find数据结构：

In this programming problem and the next you'll code up the clustering algorithm from lecture for computing a max-spacing k-clustering. Download the text file here. This file describes a distance function (equivalently, a complete graph with edge costs). It has the following format:
[number_of_nodes]
[edge 1 node 1] [edge 1 node 2] [edge 1 cost]
[edge 2 node 1] [edge 2 node 2] [edge 2 cost]
...
There is one edge (i,j) for each choice of 1≤i<j≤n, where n is the number of nodes. For example, the third line of the file is "1 3 5250", indicating that the distance between nodes 1 and 3 (equivalently, the cost of the edge (1,3)) is 5250. You can assume that distances are positive, but you should NOT assume that they are distinct.


Your task in this problem is to run the clustering algorithm from lecture on this data set, where the target number k of clusters is set to 4. What is the maximum spacing of a 4-clustering?

#include <iostream>#include <fstream>#include <sstream>#include <vector>#include <algorithm>using namespace std;typedef struct{    int nodes[2];    int cost;} Edge;bool EdgeCompare(Edge e1, Edge e2) {return (e1.cost < e2.cost);} // ------------ Union-find ------------typedef struct{    int parent;    int rank; } UnionNode;void IniUnionFind(UnionNode *nodes, int n){    for(int i=0; i<n; i++)    {        nodes[i].parent = i;        nodes[i].rank = 0;     }}int Find(UnionNode *nodes, int n, int th){    //if(nodes[th].parent == th)    //       return th;     //else    //    return Find(nodes, n, nodes[th].parent);    if(nodes[th].parent != th)        nodes[th].parent = Find(nodes, n, nodes[th].parent);    return nodes[th].parent; }void Union(UnionNode *nodes, int n, int th1, int th2){    int root1 = Find(nodes, n, th1);    int root2 = Find(nodes, n, th2);    if(root1 == root2)        return;    if (nodes[root1].rank < nodes[root2].rank)        nodes[root1].parent = root2;    else if (nodes[root1].rank > nodes[root2].rank)        nodes[root2].parent = root1;    else{nodes[root2].parent = root1;nodes[root1].rank += 1; }       }// --------------------------------------int ClusterMaxSpace(vector<Edge>& E, int nV, int k){    // ---------- Sort --------    sort(E.begin(), E.end(), EdgeCompare);    // ---------- Initial ----------    UnionNode *nodes = new UnionNode[nV];    IniUnionFind(nodes, nV);    // --------- Cluster -----------    int maxspace = 0;int nc = nV;int root1;int root2;int i = 0;while(nc != k)    {        root1 = Find(nodes, nV, E[i].nodes[0]);        root2 = Find(nodes, nV, E[i].nodes[1]);        if (root1 != root2)        {            maxspace = E[i].cost;             nc--;         }                Union(nodes, nV, E[i].nodes[0], E[i].nodes[1]);i++;    }delete nodes;    return maxspace;}int main(){    ifstream infile;    infile.open("clustering1.txt");        // ------------------------    string line;    stringstream ss;    getline(infile, line);    ss << line;    int nV;    ss >> nV;    // --------- Initialize ----------    vector<Edge> E;    ss.clear();    line.clear();    int n = 0;    Edge e;    while(getline(infile, line))    {        ss << line;                ss >> e.nodes[0];        ss >> e.nodes[1];e.nodes[0]--;e.nodes[1]--;        ss >> e.cost;                E.push_back(e);                n++;        ss.clear();        line.clear();    }infile.close();// --------------------------int k = 4;    int maxspace = ClusterMaxSpace(E, nV, k-1);     cout << maxspace << endl;    return 0;}

参考：

【1】并查集 http://zh.wikipedia.org/wiki/%E5%B9%B6%E6%9F%A5%E9%9B%86