poj_2031

来源：互联网发布：海康威视设备域名注册编辑：程序博客网时间：2024/06/08 03:44

源地址：http://poj.org/problem?id=2031

题目大意是有n个球，下面有n行，分别是每个球的x,y,z坐标和半径，叫你求将所有球连起来所需的最短长度。

两个球之间的距离被定义为求表面之间的距离，即x1*x2+y1*y2+z1*z2-r1-r2，如果两个球之间表面有接触，那么它们之间的距离就为0.题目求的其实就是最小生成树，知道题意了就不难。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64#define pi acos(-1.0);int n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;double mp[110][110];struct Node{double x,y,z;double r;}a[110];double getDis(Node A,Node B){return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+(A.z-B.z)*(A.z-B.z))-A.r-B.r;}void Prim(){int vis[110];memset(vis,0,sizeof vis);double dis[110];for(int i=0;i<n;i++)dis[i] = mp[0][i];dis[0] = 0;double ans = 0;for(int i=0;i<n;i++){int flag = -1;double tmp = INF;for(int j=0;j<n;j++)if(!vis[j]&&tmp>dis[j])tmp = dis[flag=j];if(flag == -1) break;vis[flag] = 1;ans += tmp;for(int j=0;j<n;j++)if(!vis[j] && mp[flag][j]<dis[j])dis[j] = mp[flag][j];}printf("%.3f\n",ans);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    while(sf(n)!=EOF && n){    for(int i=0;i<=n;i++)    for(int j=0;j<=n;j++)    mp[i][j] = (i==j?0:INF);    for(int i=0;i<n;i++)    scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);    for(int i=0;i<n;i++){    for(int j=i+1;j<n;j++){    double dis = getDis(a[i],a[j]);    if(dis<=0)    mp[i][j] = mp[j][i] = 0;    else    mp[i][j] = mp[j][i] = dis;    }    }    Prim();    }return 0;}

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