<OJ_Sicily>Single-link Clustering

Description

Given n nodes in a two-dimensional space, we want to use single-link custering method to findk clusters. This is equivalent to finding an MST (Minimum spanning tree) of these nodes and deletingk-1 longest edges.

Your job is to output the length of the (k-1)-th longest edges of the MST.

Input

 There are multiple cases. For each case, the first line includesn andk (2<=k<=n<=100). The following n lines give the coordinates ofn nodes. You may use Euclidean distance to measure the distance between two nodes.

Output

 For each case, output the length of the (k-1)-th longest edges. The precision is set to 2 digits after the decimal point.

题目解释：对于n个二维空间点进行聚类，将n个二维空间点分成k类。这道题转化的思想就是，先生成最小生成树，然后删除k-1个最长边

输入：第一行为节点数n以及最终分类的类别数k（2<= k <= 100)，后面n行，每一行包括两个数表示二维空间的点

输出：第k-1长的边的长度

#include <iostream>#include <algorithm>#include <math.h>#include <vector>#include <iomanip>using namespace std;vector<double> lowc;const int max_vertexs = 100;double g[max_vertexs][max_vertexs];int father[max_vertexs];typedef struct point{    int x;    int y;}P;float CalDistance(point a, point b){    return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));}void get_prim(double a[max_vertexs][max_vertexs], int vcount, int f[max_vertexs]){    }int main(int argc, const char * argv[]) {    int n,class_num;    while (cin>> n >> class_num) {        if (n == 0) break;        int x,y;        vector<point> v;        v.clear();        for (int i = 0; i < n; i++) {            cin >> x >> y;            point p1;            p1.x = x;            p1.y = y;            v.push_back(p1);        }        for (int i = 0; i < n; i ++) {            for (int j = i+1; j < n; j ++) {                g[i][j] = g[j][i] = CalDistance(v[i], v[j]);            }        }        lowc.clear();        // 使用prim算法求解最小生成树        int i, j, k;        double lowcost[max_vertexs];        int closet[max_vertexs];        int used[max_vertexs];        for (int i = 0; i < n; i ++) {            lowcost[i] = g[0][i];            closet[i] = 0;            used[i] = 0;            father[i] = -1;        }        used[0] = 1;        for (i = 1; i < n; i++) {            j = 0;            while (used[j]) {                j++;            }            for (k = 0; k < n; k ++) {                if (!used[k] && (lowcost[k] < lowcost[j])) {                    j = k;                }            }            father[j] = closet[j];            lowc.push_back(lowcost[j]);            used[j] = 1;            for (k = 0; k < n; k ++) {                if (!used[k] && (g[j][k] < lowcost[k])) {                    lowcost[k] = g[j][k];                    closet[k] = j;                }            }        }        sort(lowc.begin(), lowc.end());        cout <<  setprecision(2) << setiosflags(ios::fixed)<< lowc[n-class_num] << endl;            }    return 0;}

后记：

实现最小生成树的方法一般包括prim算法和Kruskal算法，这里使用的是prim算法来实现的。

代码新手，欢迎各位大神提出宝贵的意见和建议