POJ 2420:A Star not a Tree?
来源:互联网 发布:华为查看mac表ip 编辑:程序博客网 时间:2024/06/06 00:59
原文链接:https://www.dreamwings.cn/poj2420/2838.html
A Star not a Tree?
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5788 Accepted: 2730
Description
Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.
Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn't figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.
Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won't move his computers. He wants to minimize the total length of cable he must buy.
Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn't figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.
Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won't move his computers. He wants to minimize the total length of cable he must buy.
Input
The first line of input contains a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000.
Output
Output consists of one number, the total length of the cable segments, rounded to the nearest mm.
Sample Input
40 00 1000010000 1000010000 0
Sample Output
28284
题意:
给出平面内一个点集,让你求平面内一点距离这些点距离和的最小值~
好像退货模拟问题哎~
可惜我不会
只能用爬山算法做了,爬山算法讲的是首先在平面内选一点,判断它临近的点所求得的距离和是否比它小,若小,更新点的坐标
就这样,不过它也存在很大的局限性,如在高峰处无法选择下一点应该去哪儿,只是局部最优而已。
我们可以选择自己控制循环次数,类似于随机数之类的,然后根据大数据的判断比较,然后得出最终答案
AC代码:
#include<stdio.h>#include<math.h>#include<algorithm>#include<iostream>#define eps 1e-6using namespace std;struct point{ double x; double y; void input() //输入 { cin>>x>>y; } double dis(point k) //求两点之间距离 { return sqrt((x-k.x)*(x-k.x)+(y-k.y)*(y-k.y)); }} a[105];double getsum(point k,int n) //求距离和{ double s=0; for(int i=0; i<n; i++) s+=k.dis(a[i]); return s;}int main(){ int n; while(cin>>n) { point one; one.x=one.y=0; for(int i=0; i<n; ++i) { a[i].input(); one.x+=a[i].x; one.y+=a[i].y; } one.x/=n; one.y/=n; //选择一个靠中的点 double ans=getsum(one,n); double t=10000; while(t>eps) //循环次数 { double x=0,y=0; for(int i=0; i<n; ++i) //选取临近点 { x+=(a[i].x-one.x)/one.dis(a[i]); y+=(a[i].y-one.y)/one.dis(a[i]); } point ttm; ttm.x=one.x+x*t; ttm.y=one.y+y*t; double tmp=getsum(ttm,n); if(tmp<ans) //判断大小 { ans=tmp; one.x+=x*t; one.y+=y*t; } t*=0.99; } printf("%.0f\n",ans); } return 0;}
1 0
- POJ 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- POJ 2420 A Star not a Tree?
- 【POJ 2420】A Star not a Tree?
- POJ 2420 A Star not a Tree?
- POJ 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- poj 2420 A Star not a Tree?
- POJ 2420:A Star not a Tree?
- POJ 2420 A Star not a Tree?
- 【POJ 2420】A Star not a Tree?
- POJ 2420 A Star not a Tree? 费马点,模拟退火
- poj 2420 A Star not a Tree? 模拟退火
- POJ 2420 A Star not a Tree?(二维费马点)
- poj 2420 A Star not a Tree?(模拟退火求费马点)
- php 写内容到文件,把日志写到log文件
- 【原创】10元买啤酒2个空瓶换一瓶4个瓶盖换一瓶经典算法
- leetcode---Jump Game---贪心
- 使用Python进行Android自动化测试
- 一站式linux0.11内核head.s代码段图表详解
- POJ 2420:A Star not a Tree?
- Spring02
- 设计模式—简单工厂模式
- Spring03
- protobuf 生成对应的文件
- MD5初探及简单应用
- 2015 北京区域赛 K 二叉树乱搞(数位dp 误)
- c语言翻转一个英文句子
- 深入学习java集合:HashSet<E>实现