hdu 1875(最小生成树kruskal)

/*
  Name: 最小生成树(kruskal) 
  Author: Try_86
  Date: 10/04/12 18:51
  Description: 不符合题意所述的距离不加进边集中，然后套用kruskal就可以了 
*/
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

const int M = 5050;

int p[M], sum;
struct edge {
    int a;
    int b;
    double dis;
}e[M];
struct point {
    double x;
    double y;
}po[101];

int cmp(const void *a, const void *b){
    if ((*(edge *)a).dis > (*(edge *)b).dis) return 1;
    return -1;
}

void init(int vs) {
    for (int i=1; i<=vs; ++i) p[i] = i;
    return ;
}

int find(int v) {
    if (p[v] != v) p[v] = find(p[v]);
    return p[v];
}

double join(edge e) {
    int x, y;
    x = find(e.a);
    y = find(e.b);
    if (x != y) {
        ++sum;
        p[x] = y;
        return e.dis;
    }
    return 0;
}

double kruskal(int es, int vs) {
    double ans = 0;
    init(vs);
    qsort(e, es, sizeof(edge), cmp);
    for (int i=0; i<es; ++i) {
        ans += join(e[i]);
        if (sum == vs) return ans;
    } 
    if (sum < vs) return -1;
}

int main() {
    int t;
    scanf ("%d", &t);
    while (t--) {
        int n, es;
        scanf ("%d", &n);
        for (int i=1; i<=n; ++i) scanf ("%lf%lf", &po[i].x, &po[i].y);
        if (n <= 1) {
            printf ("0.0\n");
            continue;
        }
        double dis;
        es = 0;
        for (int i=1; i<n; ++i) {
            for (int j=i+1; j<=n; ++j) {
                dis = sqrt((po[i].x-po[j].x)*(po[i].x-po[j].x)+(po[i].y-po[j].y)*(po[i].y-po[j].y));
                if (dis >= 10.0 && dis <= 1000.0) {
                    e[es].dis = dis;
                    e[es].a = i;
                    e[es].b = j;
                    ++es;
                }
            }
        }
        sum = 1;
        double ans = kruskal(es, n);
        if (ans < 0) printf ("oh!\n");
        else printf ("%.1lf\n", ans*100); 
    }
    return 0;
}