POJ 2031 (最小生成树)

来源：互联网发布：软件需求分为哪几类编辑：程序博客网时间：2024/06/06 04:19

题意是n个球求MST,球相交或者相切距离认为是0.

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;#define maxn 111struct yuan {    double x, y, z, r;}p[maxn];struct node {    int u, v;    double w;    bool operator < (const node &a) const {        return w < a.w;    }} edge[maxn*maxn];int n, m;double dis (yuan a, yuan b) {    double xx = a.x-b.x, yy = a.y-b.y, zz = a.z-b.z;    double d = sqrt (xx*xx + yy*yy + zz*zz);    if (d <= a.r+b.r)        return 0;    return d - a.r - b.r;}#define find Findint fa[maxn];int find (int x) {    return fa[x] == x ? fa[x] : fa[x] = find (fa[x]);}int main () {    //freopen ("in", "r", stdin);    while (cin >> n && n) {        m = 0;        for (int i = 1; i <= n; i++)            fa[i] = i;        for (int i = 1; i <= n; i++) {            cin >> p[i].x >> p[i].y >> p[i].z >> p[i].r;            for (int j = 1; j < i; j++) {                edge[m].u = i, edge[m].v = j, edge[m++].w = dis (p[i], p[j]);            }        }        sort (edge, edge+m);        double ans = 0.0;        int cnt = 0;        for (int i = 0; i < m; i++) {            int p1 = find (edge[i].u), p2 = find (edge[i].v);            if (p1 != p2) {                fa[p1] = p2;                ans += edge[i].w;                cnt++;            }            if (cnt == n-1)                break;        }        printf ("%.3f\n", ans);    }    return 0;}

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