Building a Space Station Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4400 Accepted: 2255 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are ex…

Building a Space Station POJ - 2031 You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task. The space station is ma…

http://poj.org/problem?id=2031 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task.  The space stati…

题目传送门 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_2…

Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task.The space station is made up with a number of un…

http://poj.org/problem?id=2031 题意 给出三维坐标系下的n个球体,求把它们联通的最小代价. 分析 最小生成树加上一点计算几何.建图,若两球体原本有接触,则边权为0:否则边权为它们球心的距离-两者半径之和.这样来跑Prim就ok了.注意精度. #include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #in…

链接: http://poj.org/problem?id=2031 Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6011   Accepted: 2989 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the stati…

http://poj.org/problem?id=2421 这个题和poj1258是一样的,只要在1258的基础上那么几行代码,就可以A,水. 题意:还是n连通问题,和1258不同的就是这个还有几条路在之前就已经连通了的,所以不需要再去连. #include <stdio.h> #include <string.h> #define inf 100009 ]; ][],dis[],ans,n; int prim() { ;i<=n;i++) dis[i]=inf;dis[]…

题目链接:http://poj.org/problem?id=1251 字符用%s好了,方便一点. #include <stdio.h> #include <string.h> #define INF 0x3f3f3f3f ][]; ]; ]; int n; int Prim() { memset(vis,false,sizeof(vis)); ; i<=n; i++) dis[i] = INF; ; dis[] = ; ; i<=n; i++) { ; ; j<…

这个题要交c++, 因为prime的返回值错了,改了一会 题目:http://poj.org/problem?id=2031 题意:就是给出三维坐标系上的一些球的球心坐标和其半径,搭建通路,使得他们能够相互连通.如果两个球有重叠的部分则算为已连通,无需再搭桥.求搭建通路的最小费用(费用就是边权,就是两个球面之间的距离 思路:求球之间的距离, 距离小于半径 说明联通 , 距离为0. #include<iostream> #include<cstdio> #include<cst…