Run Away
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 6482   Accepted: 1993

Description

One of the traps we will encounter in the Pyramid is located in the Large Room. A lot of small holes are drilled into the floor. They look completely harmless at the first sight. But when activated, they start to throw out very hot java, uh ... pardon, lava. Unfortunately, all known paths to the Center Room (where the Sarcophagus is) contain a trigger that activates the trap. The ACM were not able to avoid that. But they have carefully monitored the positions of all the holes. So it is important to find the place in the Large Room that has the maximal distance from all the holes. This place is the safest in the entire room and the archaeologist has to hide there.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing three integers X, Y, M separated by space. The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= M <= 1000. The numbers X and Yindicate the dimensions of the Large Room which has a rectangular shape. The number M stands for the number of holes. Then exactly M lines follow, each containing two integer numbers Ui and Vi (0 <= Ui <= X, 0 <= Vi <= Y) indicating the coordinates of one hole. There may be several holes at the same position.

Output

Print exactly one line for each test case. The line should contain the sentence "The safest point is (P, Q)." where P and Qare the coordinates of the point in the room that has the maximum distance from the nearest hole, rounded to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1).

Sample Input

3
1000 50 1
10 10
100 100 4
10 10
10 90
90 10
90 90
3000 3000 4
1200 85
63 2500
2700 2650
2990 100

Sample Output

The safest point is (1000.0, 50.0).
The safest point is (50.0, 50.0).
The safest point is (1433.0, 1669.8). 思路:直接按照题解做的,首先列举30个随机点,然后分别进行短距离的随机尝试
#include <cstdio>
#include <ctime>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
const double Pi=acos(-1.0);
const double eps=1e-3;
const double inf=1e20;
double px[31],py[31],d[31],x[1111],y[1111];
double caldis(double x1,double y1,double x2,double y2){
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main(){
int T;
scanf("%d",&T);
srand(9876543);
while(T--){
int a,b,m;
scanf("%d%d%d",&a,&b,&m);
for(int i=0;i<m;i++){
scanf("%lf%lf",x+i,y+i);
}
for(int i=0;i<30;i++){
px[i]=(double)(rand()%1000+1)/1000.000*a;
py[i]=(double)(rand()%1000+1)/1000.000*b;
d[i]=inf;
for(int j=0;j<m;j++)d[i]=min(d[i],caldis(px[i],py[i],x[j],y[j]));
}
double delta=double(max(a,b))/(sqrt(1.0*m));
while(delta>eps){
for(int i=0;i<30;i++){
double tx=px[i],ty=py[i];
for(int j=0;j<30;j++){
double theta=(double)(rand()%1000+1)/1000.000*10*Pi;
double dx=delta*cos(theta);
double dy=delta*sin(theta);
tx+=dx;ty+=dy;
if(tx<0||tx>a||ty<0||ty>b){tx-=dx;ty-=dy;continue;}
double td=inf;
for(int k=0;k<m;k++)td=min(td,caldis(tx,ty,x[k],y[k]));
if(td>d[i]){
d[i]=td;px[i]=tx;py[i]=ty;
}
tx-=dx;ty-=dy;
}
}
delta*=0.9;
}
double ans=0;int ind=0;
for(int i=0;i<30;i++){
if(d[i]>ans){
ind=i;
ans=d[i];
}
}
printf("The safest point is (%.1f, %.1f).\n",px[ind],py[ind]);
}
return 0;
}

  

poj 1379 Run Away 模拟退火 难度:1的更多相关文章

  1. POJ.1379.Run Away(模拟退火)

    题目链接 POJ输出不能用%lf! mmp从4:30改到6:00,把4:30交的一改输出也过了. 于是就有了两份代码.. //392K 500MS //用两点构成的矩形更新,就不需要管边界了 #inc ...

  2. POJ 1379 Run Away 【基础模拟退火】

    题意:找出一点,距离所有所有点的最短距离最大 二维平面内模拟退火即可,同样这题用最小圆覆盖也是可以的. Source Code: //#pragma comment(linker, "/ST ...

  3. PKU 1379 Run Away(模拟退火算法)

    题目大意:原题链接 给出指定的区域,以及平面内的点集,求出一个该区域内一个点的坐标到点集中所有点的最小距离最大. 解题思路:一开始想到用随机化算法解决,但是不知道如何实现.最后看了题解才知道原来是要用 ...

  4. POJ 1379 Run Away

    题意:有n个陷阱,在X,Y范围内要求出一个点使得这个点到陷阱的最小距离最大. 思路:模拟退火,随机撒入40个点,然后模拟退火随机化移动. (这题poj坑爹,加了srand(time(NULL))不能交 ...

  5. POJ 1379 模拟退火

    模拟退火算法,很久之前就写过一篇文章了.双倍经验题(POJ 2420) 题意: 在一个矩形区域内,求一个点的距离到所有点的距离最短的那个,最大. 这个题意,很像二分定义,但是毫无思路,也不能暴力枚举, ...

  6. POJ 1379 (随机算法)模拟退火

    题目大意: 给定一堆点,找到一个点的位置使这个点到所有点中的最小距离最大 这里数据范围很小,精度要求也不高,我们这里可以利用模拟退火的方法,随机找到下一个点,如果下一个点比当前点优秀就更新当前点 参考 ...

  7. poj-1379 Run Away(模拟退火算法)

    题目链接: Run Away Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7982   Accepted: 2391 De ...

  8. 【BZOJ1844/2210】Pku1379 Run Away 模拟退火

    [BZOJ1844/2210]Pku1379 Run Away 题意:矩形区域中有一堆点,求矩形中一个位置使得它到所有点的距离的最小值最大. 题解:模拟退火的裸题,再调调调调调参就行了~ #inclu ...

  9. poj 2069 Super Star —— 模拟退火

    题目:http://poj.org/problem?id=2069 仍是随机地模拟退火,然而却WA了: 看看网上的题解,都是另一种做法——向距离最远的点靠近: 于是也改成那样,竟然真的A了...感觉这 ...

随机推荐

  1. MySQL 温故知心(三)

    MySQL锁概述 相对其他数据库而言,MySQL的锁机制比较简单,其最显著的特点是不同的存储引擎支持不同的锁机制.比如,MyISAM和MEMORY存储引擎采用的是表级锁(table-level loc ...

  2. 46. Permutations (全排列)

    Given a collection of distinct numbers, return all possible permutations. For example,[1,2,3] have t ...

  3. 论文笔记:OverFeat: Integrated Recognition, Localization and Detection using Convolutional Networks

    2014 ICLR 纽约大学 LeCun团队 Pierre Sermanet, David Eigen, Xiang Zhang, Michael Mathieu, Rob Fergus, Yann ...

  4. Eclipase .自动提示

    打开eclipse→Windows→Preferences→Java→Editor→Content Assist 修改Auto Activation triggers for java的值为 .abc ...

  5. Linux下修改时间

    修改linux的时间可以使用date指令 date命令的功能是显示和设置系统日期和时间. 输入date 查看目前系统时间. 修改时间需要 date -功能字符 修改内容 命令中各选项的含义分别为: - ...

  6. ansible一些基本操作

    一.介绍 特性 (1).no agents:不需要在被管控主机上安装任何客户端: (2).no server:无服务器端,使用时直接运行命令即可: (3).modules in any languag ...

  7. asp.net Mvc 使用uploadify 上传文件 HTTP 302 Error

    CSHTML代码 @{ if (Request.Cookies[FormsAuthentication.FormsCookieName] != null) { <input type=" ...

  8. STM32-串行SPI nor

    源:FLASH 存储学习-串行SPI nor 1.1 SST25VF080B简介1.1.1 主要特性 关键点:容量.速度(时钟速度.读写速度).功耗. l 容量:8MBit: l 最高SPI时钟频率: ...

  9. 【c++ primer, 5e】设计Sales_data类 & 定义改进的Sales_data类

    [设计Sales_data类] 1.考虑Sales_data类应该有那些接口. isbn.combine.add.read.print... 2.考虑如何使用这些接口. Sales_data tota ...

  10. jQuery :gt 选择器 jQuery :lt 选择器

    选择前 3 个之后的所有 <tr> 元素: $("tr:gt(2)"); 选择前 2 个 <tr> 元素: $("tr:lt(2)");