Prim POJ 2031 Building a Space Station-白红宇

Prim POJ 2031 Building a Space Station

阅读量：5331 次

发布时间：2019-06-14

本文共 2460 字，大约阅读时间需要 8 分钟。

题意：给出n个三维空间的球体，球体是以圆心坐标+半径来表示的，要求在球面上建桥使所有的球联通，求联通所建桥的最小长度。

分析：若两点距离大于两半径和的长度，那么距离就是两点距离 - 半径和，否则为0，Prim写错了，算法没有完全理解

/************************************************* Author        :Running_Time* Created Time  :2015/10/25 12:00:48* File Name     :POJ_2031.cpp ************************************************/#include 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 1e2 + 10;const int E = N * N;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const double EPS = 1e-10;bool vis[N];double d[N];int head[N];int n, m, e;int dcmp(double x) { //三态函数，减少精度问题 if (fabs (x) < EPS) return 0; else return x < 0 ? -1 : 1; } struct Point { double x, y, z, a; Point () {} Point (double x, double y, double z, double a) : x (x), y (y), z (z), a (a) {} Point operator - (const Point &r) const { //向量减法 return Point (x - r.x, y - r.y, z - r.z, 0); } };typedef Point Vector; //向量的定义 Point read_point(void) { //点的读入 double x, y, z, r; scanf ("%lf%lf%lf%lf", &x, &y, &z, &r); return Point (x, y, z, r); } double dot(Vector A, Vector B) { //向量点积 return A.x * B.x + A.y * B.y + A.z * B.z; } double length(Vector A) { //向量长度，点积 return sqrt (dot (A, A)); } struct Edge { int v, nex; double w; Edge () {} Edge (int v, double w, int nex) : v (v), w (w), nex (nex) {} bool operator < (const Edge &r) const { return w > r.w; }}edge[E];void init(void) { memset (head, -1, sizeof (head)); e = 0;}void add_edge(int u, int v, double w) { edge[e] = Edge (v, w, head[u]); head[u] = e++;}double Prim(int s) { memset (vis, false, sizeof (vis)); for (int i=0; i
                      
                        Q; for (int i=head[s]; ~i; i=edge[i].nex) { int v = edge[i].v; double w = edge[i].w; if (d[v] > w) { d[v] = w; Q.push (Edge (v, d[v], 0)); } } vis[s] = true; d[s] = 0; double ret = 0; while (!Q.empty ()) { int u = Q.top ().v; Q.pop (); if (vis[u]) continue; vis[u] = true; ret += d[u]; for (int i=head[u]; ~i; i=edge[i].nex) { int v = edge[i].v; double w = edge[i].w; if (!vis[v] && d[v] > w) { d[v] = w; Q.push (Edge (v, d[v], 0)); } } } return ret;}Point p[N];int main(void) { while (scanf ("%d", &n) == 1) { if (!n) break; for (int i=0; i

转载于:https://www.cnblogs.com/Running-Time/p/4908682.html

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