Run Away

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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).
 
 #include <iostream>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <algorithm>
using namespace std;
#define N 50
#define M 10
struct point
{
double x,y,mina;
};
point p[],s,tri[];
int n;
double dist(point a,point b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
double getmina(point tem)
{
double mina=1e9,t;
for(int i=; i<n; i++)
{
t=dist(tem,p[i]);
if(t<mina)
mina=t;
}
return mina;
}
void Simulated_Annealing()
{
int i,j;
for(i=; i<N; i++)
{
tri[i].x=((rand()%)+1.0)/1000.0*s.x;
tri[i].y=((rand()%)+1.0)/1000.0*s.y;
tri[i].mina=getmina(tri[i]);
}
double temper=s.x+s.y,dx,dy;
point tmp;
while(temper>0.001)
{
for(i=; i<N; i++)
{
for(j=; j<M; j++)
{
dx=((rand()%)+1.0)/1000.0*temper;
dy=sqrt(temper*temper-dx*dx);
if (rand()&) dx*=-;
if (rand()&) dy*=-;
tmp.x=tri[i].x+dx;
tmp.y=tri[i].y+dy;
if (tmp.x>= && tmp.x<=s.x && tmp.y>= && tmp.y<=s.y)
{
tmp.mina=getmina(tmp);
if(tmp.mina>tri[i].mina)
{
tri[i]=tmp;
}
}
}
}
temper*=0.6;
}
int mini=;
for(i=;i<N;i++)
if(tri[mini].mina<tri[i].mina)
mini=i;
printf("The safest point is (%.1lf, %.1lf).\n",tri[mini].x,tri[mini].y);
}
int main()
{
srand((unsigned int)(time(NULL)));
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%d",&s.x,&s.y,&n);
for(i=; i<n; i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
Simulated_Annealing();
}
}

Run Away 模拟退火的更多相关文章

  1. 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 ...

  2. 【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 ...

  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 模拟退火 难度:1

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

  5. 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 ...

  6. 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 ...

  7. 模拟退火算法(run away poj1379)

    http://poj.org/problem?id=1379 Run Away Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: ...

  8. 【模拟退火】poj1379 Run Away

    题意:平面上找一个点,使得其到给定的n个点的距离的最小值最大. 模拟退火看这篇:http://www.cnblogs.com/autsky-jadek/p/7524208.html 这题稍有不同之处仅 ...

  9. hdu3932 模拟退火

    模拟退火绝对是从OI--ACM以来接触过的所有算法里面最黑科技的orz 题意:地上有一堆hole,要找一个点,使得(距离该点最远的hole的距离)最小. sol:本来想套昨天的模拟退火模板,初值(0, ...

随机推荐

  1. selenium 对chrome浏览器操作

    参照http://www.testwo.com/blog/6931博客内容 1.下载ChromeDriver驱动包(下载地址: http://chromedriver.storage.googleap ...

  2. PHP常量定义define与const

    一.const PHP5.3以前,const只能在类内部声明变量,5.3+允许在外部声明变量,但还不能使用常量计算! const ONE = 1; const WORD = 'hello world' ...

  3. Microsoft .Net Remoting系列专题之一:.Net Remoting基础篇

    Microsoft .Net Remoting系列专题之一 一.Remoting基础 什么是Remoting,简而言之,我们可以将其看作是一种分布式处理方式.从微软的产品角度来看,可以说Remotin ...

  4. selenium元素定位不到之iframe

    我们在使用selenium的18中定位方式的时候,有时会遇到定位不上的问题,今天我们就来说说导致定位不上的其中一个原因---iframe 问题描述:通过firebug查询到相应元素的id或name等, ...

  5. 最新城市二级联动json(2017-09)

    { '安徽': [ '合肥', '芜湖', '蚌埠', '淮南', '马鞍山', '淮北', '铜陵', '安庆', '黄山', '阜阳', '宿州', '滁州', '六安', '宣城', '池州', ...

  6. [转载]Python实现浏览器自动化操作

    原文地址:Python实现浏览器自动化操作作者:rayment   最近在研究网站自动登录的问题,涉及到需要实现浏览器自动化操作,网上有不少介绍,例如使用pamie,但是只是支持IE,而且项目也较久没 ...

  7. 转:【Java集合源码剖析】Java集合框架

    转载轻注明出处:http://blog.csdn.net/ns_code/article/details/35564663   Java集合工具包位于Java.util包下,包含了很多常用的数据结构, ...

  8. The Last

    第八次课程作业 感慨 没想到这就最后一次课程作业了,还以为会跟我到大学毕业呢.既然是最后一次就说说心里话.起初收到做博客作业的消息还觉得蛮有新意的(因为第一次作业不难),后来不断的作业涌现出来了,还都 ...

  9. 201521123011《Java程序设计》第14周学习总结

    1. 本周学习总结 1.1 以你喜欢的方式(思维导图或其他)归纳总结多数据库相关内容. MySql数据库简单操作 1.启动与退出(quit或exit ) 操作 显示所有数据库: show databa ...

  10. 控制结构(3) 状态机(state machine)

    // 上一篇:卫语句(guard clause) // 下一篇:局部化(localization) 基于语言提供的基本控制结构,更好地组织和表达程序,需要良好的控制结构. 前情回顾 上次分析了guar ...