POJ.1379.Run Away(模拟退火)

weixin_30852419

于 2018-04-28 19:00:00 发布

阅读量108

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/SovietPower/p/8969287.html

本文介绍了一种使用随机搜索算法解决POJ题目中寻找最安全点的问题，通过不断迭代和缩小搜索范围来找到距离所有危险点最远的位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
#include <cmath>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
#define Rate (0.9)
#define Rand() (1.0*((rand()<<1)-RAND_MAX)/RAND_MAX)
const int N=1005;

int n,X,Y;
double DIS,ans_dis[20];
struct Node{
    double x,y;
    Node() {}
    Node(double x,double y):x(x),y(y) {}
}p[N],ans[30];

inline int read()
{
    int now=0;register char c=gc();
    for(;!isdigit(c);c=gc());
    for(;isdigit(c);now=now*10+c-'0',c=gc());
    return now;
}
inline double Sqr(double x){
    return x*x;
}
inline double Dis(double x,double y,int i){
    return sqrt(Sqr(x-p[i].x)+Sqr(y-p[i].y));
}
double Calc(double x,double y)
{
    double res=DIS;
    for(int i=1; i<=n; ++i)
        res=std::min(res,Dis(x,y,i));
    return res;
}
void Print(){
    for(int i=1; i<=15; ++i) printf("%d:%.5lf %.5lf %.5lf   %d %d\n",i,ans[i].x,ans[i].y,ans_dis[i],X,Y);
}
void Solve()
{
    DIS=sqrt(X*X+Y*Y);
    const int tot=15;
    for(int i=1; i<=tot; ++i)
        ans[i].x=1.0*rand()/RAND_MAX*X,ans[i].y=1.0*rand()/RAND_MAX*Y,ans_dis[i]=Calc(ans[i].x,ans[i].y);
    double T=DIS,xx,yy,dis;//this...
    while(T>0.01)
    {
        for(int i=1; i<=tot; ++i)
            for(int j=1; j<=30; ++j)
            {
                xx=ans[i].x+T*Rand(), yy=ans[i].y+T*Rand();
                if(xx<0||xx>X||yy<0||yy>Y) continue;//!
                dis=Calc(xx,yy);
                if(dis>ans_dis[i]) ans[i]=Node(xx,yy),ans_dis[i]=dis;
            }
        T*=Rate;
    }
    int res=1;
    for(int i=2; i<=tot; ++i)
        if(ans_dis[i]>ans_dis[res]) res=i;
    printf("The safest point is (%.1f, %.1f).\n",ans[res].x,ans[res].y);
}

int main()
{
    srand(20180428);
    int T=read();
    while(T--)
    {
        X=read(),Y=read(),n=read();
        for(int i=1; i<=n; ++i) p[i].x=read(),p[i].y=read();
        Solve();
    }
    return 0;
}

另一种写法：

//392K  1188MS
#include <cmath>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
#define Rate (0.97)
//#define Rand() (1.0*((rand()<<1)-RAND_MAX)/RAND_MAX)
#define Rand() (1.0*rand()/RAND_MAX)

const int N=1005;

int n,X,Y;
double DIS,ans_dis[30];
struct Node{
    double x,y;
    Node() {}
    Node(double x,double y):x(x),y(y) {}
}p[N],ans[30];

inline int read()
{
    int now=0;register char c=gc();
    for(;!isdigit(c);c=gc());
    for(;isdigit(c);now=now*10+c-'0',c=gc());
    return now;
}
inline double Sqr(double x){
    return x*x;
}
inline double Dis(double x,double y,int i){
    return sqrt(Sqr(x-p[i].x)+Sqr(y-p[i].y));
}
double Calc(double x,double y)
{
    double res=1e9;//DIS;
    for(int i=1; i<=n; ++i)
        res=std::min(res,Dis(x,y,i));
    return res;
}
void Rand_Point(double &x,double &y,Node a,Node b){
    x=a.x+(b.x-a.x)*Rand(), y=a.y+(b.y-a.y)*Rand();
}
void Solve()
{
    DIS=sqrt(X*X+Y*Y);
    const int tot=15;
    ans[1]=Node(0,0), ans[2]=Node(X,Y), ans[3]=Node(X,0), ans[4]=Node(0,Y);//
    for(int i=5; i<=tot; ++i)
        Rand_Point(ans[i].x,ans[i].y,ans[1],ans[2]);
//      ans[i].x=Rand()*X,ans[i].y=Rand()*Y;
    for(int i=1; i<=tot; ++i) ans_dis[i]=Calc(ans[i].x,ans[i].y);
    double T=std::min(X,Y),xx,yy,dis;//this...
    while(T>0.01)
    {
        for(int i=1; i<=tot; ++i)
            for(int j=1; j<=20; ++j)
            {
                Rand_Point(xx,yy,Node(std::max(0.0,ans[i].x-T),std::max(0.0,ans[i].y-T)),Node(std::min(1.0*X,ans[i].x+T),std::min(1.0*Y,ans[i].y+T)));
                dis=Calc(xx,yy);
                if(dis>ans_dis[i]) ans[i]=Node(xx,yy),ans_dis[i]=dis;
            }
        T*=Rate;
    }
    int res=1;
    for(int i=2; i<=tot; ++i)
        if(ans_dis[i]>ans_dis[res]) res=i;
    printf("The safest point is (%.1f, %.1f).\n",ans[res].x,ans[res].y);
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
//  freopen("1.out","w",stdout);
#endif

    srand(20180428);
    int T=read();
    while(T--)
    {
        X=read(),Y=read(),n=read();
        for(int i=1; i<=n; ++i) p[i].x=read(),p[i].y=read();
        Solve();
    }
    return 0;
}

转载于:https://www.cnblogs.com/SovietPower/p/8969287.html