C

Think:
1知识点：最小生成树_Prim算法+浮点数坐标建图
2题意分析：给定n个点的空间坐标和半径，将n个点直接或间接连通，连通可认为:
1>两个坐标(作为球理解)重合或覆盖
2>建立通道将两个坐标(作为球理解)连接，通道长度计算应为表面到表面的直线长度
3优化思路：题目中自己使用浮点数建图，后续优化可考虑使用实数建图，初始坐标输入的时候将浮点数转化为实数(本题目乘以1000即可)，需要注意的是在求坐标距离的时候注意平方之后求和结果不要超出int范围，可考虑直接强制类型转换为double，进而直接开平方得到浮点数，进而再转化为实数(本题目乘以1000即可)，最终求出连通后的最短总长度再强制类型转换之后再转化为浮点数(除以1000)，认为即可一定程度降低误差

vjudge题目链接

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int inf = 0x3f3f3f3f;const double zero = 0.001;struct Node{    double x;    double y;    double z;    double r;}red[114];double e[114][114], dis[114];int vis[114];double Dist(int u, int v);void Prim(int n);int main(){    int n, i, j;    while(scanf("%d", &n) && n){        for(i = 1; i <= n; i++){            scanf("%lf %lf %lf %lf", &red[i].x, &red[i].y, &red[i].z, &red[i].r);        }        for(i = 1; i <= n; i++){            e[i][i] = inf;            for(j = i+1; j <= n; j++){                e[i][j] = e[j][i] = Dist(i, j);            }        }        Prim(n);    }    return 0;}double Dist(int u, int v){    double dx = red[u].x - red[v].x;    double dy = red[u].y - red[v].y;    double dz = red[u].z - red[v].z;    double R = dx*dx + dy*dy + dz*dz;    double dr = sqrt(R);    dr = dr - red[u].r - red[v].r;    if(dr < zero)        return 0.000;    else        return dr;}void Prim(int n){    int i, v, num;    double miv, sum;    memset(vis, 0, sizeof(vis));    for(i = 1; i <= n; i++)        dis[i] = e[1][i];    vis[1] = 1, dis[1] = 0.000, num = 1, sum = 0.000;    while(num < n){        miv = inf;        for(i = 1; i <= n; i++){            if(!vis[i] && miv - dis[i] > zero){                miv = dis[i], v = i;            }        }        vis[v] = 1, num++, sum += dis[v];        for(i = 1; i <= n; i++){            if(!vis[i] && dis[i] - e[v][i] > zero)                dis[i] = e[v][i];        }    }    printf("%.3lf\n", sum);}