[2016-04-14][POJ][203][Building a Space Station]

• 时间:2016-04-14 21:43:30 星期四

• 题目编号:[2016-04-14][POJ][203][Building a Space Station]

• 题目大意:给定n个球体,每个球体可能重合,可能包含,可能分离,问把每个球体连接起来(重合和包含看做已经连接),至少需要多长的路

• 分析:最小生成树,边权为 max(0,disij−ri−rj)$max(0,di{s}_{ij}-r_i-r_j)$,即重合和内含,边权为0

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 100 + 10;
double g[maxn][maxn],lowc[maxn];
int vis[maxn];
struct Point{
        double x,y,z,r;
        Point(double a = 0,double b = 0,double c = 0,double d = 0):x(a),y(b),z(c),r(d){}
}p[maxn];
double dtmp(double a,double b){
        return (a - b)*(a - b);
}
double dis(int a,int b){
        return sqrt(dtmp(p[a].x,p[b].x) + dtmp(p[a].y,p[b].y) + dtmp(p[a].z,p[b].z));
}
double prim(int n){
        memset(vis,0,sizeof(vis));
        vis[0] = 1;
        for(int i = 1 ; i < n ; ++i){
                lowc[i] = g[0][i];
        }
        double ans = 0;
        for(int i = 1 ; i < n ; ++i){
                double minc = 999999999;
                int p = -1;
                for(int j = 0 ; j < n ; ++j){
                        if(!vis[j] && minc > lowc[j]){
                                minc = lowc[j];
                                p = j;
                        }
                }
                if( minc == 999999999)        return -1;
                ans += minc;
                vis[p] = 1;
                for(int j = 0 ; j <  n ; ++j){
                        if(!vis[j] && lowc[j] > g[p][j]){
                                lowc[j] = g[p][j];
                        }
                }
        }
        return ans;
}
int main(){
        int n;
        while(~scanf("%d",&n) && n){
                double a,b,c,d,tmp;
                for(int i = 0 ; i < n ; ++i){
                        scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
                        p[i] = Point(a,b,c,d);
                }               
                for(int i = 0 ; i < n ; ++i){
                        for(int j = 0 ; j < n ; ++j){
                                tmp = dis(i,j) - p[i].r - p[j].r;
                                g[i][j] = max(double(0.0),tmp);
                        }
                }
                printf("%.3f\n",prim(n));
        }
        return 0;
}

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