Bzoj 1821 [Jsoi2010]group Tribe Division (Kruskal)

Solution: At first didn't think how to write, but slowly feel that the distance between the same tribe is as short as possible, and then, like the picture in the title, the distance between the two tribes will be very long, this length is actually by the tribe inside a person to go out and another tribe of the distance. We run the smallest spanning tree directly, so that the distance of the people of their tribe is as short as possible, then this spanning tree the longest several sides inside, directly output the penultimate k-1 bar.

The code is as follows:

#include <iostream> #include <cstdio> #include <vector> #include <queue> #include <utility > #include <stack> #include <algorithm> #include <cstring> #include <string> #include <
cmath> #include <ctime> #include <set> using namespace std;
typedef pair<int, int> PII;
const int MAXN = 1e5 + 5;
const int MAXM = 1e6 + 5;
int N, K, PPP;
PII P[MAXN];

vector<double>len;
    struct Node {int u, v;
    Double C;
        BOOL operator < (CONST node& TMP) const {if (C < tmp.c) return 1;
    return 0;

}}EDGE[MAXM];
int PAR[MAXN];
    int find (int x) {int root, TMP = x;
    while (x!= par[x]) x = par[x];
    root = x;
    x = tmp;
        while (x!= par[x]) {tmp = par[x];
        PAR[X] = root;
    x = tmp;
return root;

} void Union (int x, int y) {par[y] = x;}
        void Kruskal () {for (int i = 0; i < PPP; i++) {int x = find (EDGE[I].U); InT y = find (EDGE[I].V);
            if (x!= y) {Union (x, y);
        Len.push_back (EDGE[I].C);
    Double Count (int i, int j) {int x1 = p[i].first, x2 = P[j].first;
    int y1 = p[i].second, y2 = P[j].second;
return sqrt (1.0 * (x1-x2) * (X1-X2) + (y1-y2) * (Y1-y2)); int main () {#ifndef Online_judge freopen ("D:\\in.txt", "R", stdin); #endif//Online_judge scanf ("%d%d" &
    n, &k);
        for (int i = 1, x, y; i <= n; i++) {scanf ("%d%d", &x, &y);
        P[i].first = x;
    P[i].second = y;
            for (int i = 1; I <= n; i++) {for (int j = i + 1; j <= N; j +) {Double d = Count (i, j);
            EDGE[PPP].U = i;
            EDGE[PPP].V = j;
            EDGE[PPP].C = D;
        ppp++;
    } sort (Edge, Edge + PPP);
    for (int j = 1; j < Maxn; J + +) par[j] = j;
    Kruskal ();
	printf ("%.2f\n", Len[len.size ()-K + 1]);
return 0;
 }