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;
}