POJ1379 Run Away

标签：can   exp   代码   sqrt   safe   xxx   平面   卡死   i++   

我对模拟退火的理解：https://www.cnblogs.com/AKMer/p/9580982.html

题目传送门：http://poj.org/problem?id=1379

题目意思是要求规定大小平面内某一点，该点到所有给定点的距离中最小距离最大。

类似于费马点做法……不过把统计距离和变成了求距离最小值最大。

费马点不会的可以先看看这篇博客。

BZOJ3680：吊打XXXhttps://www.cnblogs.com/AKMer/p/9588724.html

时间复杂度：\(O(能A)\)

空间复杂度：\(O(能A)\)

爬山算法代码如下：

#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;

#define sqr(a) ((a)*(a))

int n,lenx,leny;
double ansx,ansy,ans;

int read() {
    int x=0,f=1;char ch=getchar();
    for(;ch<‘0‘||ch>‘9‘;ch=getchar())if(ch==‘-‘)f=-1;
    for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())x=x*10+ch-‘0‘;
    return x*f;
}

struct point {
    double x,y;
}p[1001];

double len() {
    double x=rand()%200000-100000;
    return x/100000;
}

double dis(double x1,double y1,double x2,double y2) {
    return sqrt(sqr(x1-x2)+sqr(y1-y2));
}

double calc(double x,double y) {
    double tmp=1e18;
    for(int i=1;i<=n;i++)
        tmp=min(tmp,dis(x,y,p[i].x,p[i].y));//求最小距离最大
    if(tmp>ans) {ans=tmp;ansx=x,ansy=y;}
    return tmp;
}

void climbhill() {
    double now_x=ansx,now_y=ansy;
    for(double T=1e7;T>=1e-7;T*=0.998) {
        double nxt_x=now_x+len()*T;
        double nxt_y=now_y+len()*T;
        if(nxt_x<0||nxt_x>lenx||nxt_y<0||nxt_y>leny)continue;
        if(calc(now_x,now_y)<calc(nxt_x,nxt_y))
            now_x=nxt_x,now_y=nxt_y;
    }
}

int main() {
    srand(time(0));
    int T=read();
    while(T--) {
        lenx=read(),leny=read(),n=read();
        ans=-1e18;ansx=lenx/2;ansy=leny/2;//起始点设在平面中央
        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        climbhill();
        printf("The safest point is (%.1lf, %.1lf).\n",ansx,ansy);
    }
    return 0;
}

模拟退火代码如下：

#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;

#define sqr(x) ((x)*(x))

const double T_0=1e-7,del_T=0.998;

int lenx,leny,n;
double ansx,ansy,ans;

int read() {
    int x=0,f=1;char ch=getchar();
    for(;ch<‘0‘||ch>‘9‘;ch=getchar())if(ch==‘-‘)f=-1;
    for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())x=x*10+ch-‘0‘;
    return x*f;
}

struct point {
    double x,y;
}p[1001];

double len() {
    double x=rand()%200000-100000;
    return x/100000;
}

double dis(double x1,double y1,double x2,double y2) {
    return sqrt(sqr(x1-x2)+sqr(y1-y2));
}

double calc(double x,double y) {
    double tmp=1e18;
    for(int i=1;i<=n;i++)
        tmp=min(tmp,dis(x,y,p[i].x,p[i].y));
    if(tmp>ans) {ans=tmp;ansx=x;ansy=y;}
    return tmp;
}

void Anneal() {
    double T=1e7,now_x=ansx,now_y=ansy;
    while(T>=T_0) {
        double nxt_x=now_x+len()*T;
        double nxt_y=now_y+len()*T;
        if(nxt_x<0||nxt_x>lenx||nxt_y<0||nxt_y>leny) {
            T*=del_T;continue;//不这么写直接卡死循环了……
        }
        double tmp1=calc(now_x,now_y);
        double tmp2=calc(nxt_x,nxt_y);
        if(tmp2>tmp1||exp((tmp2-tmp1)/T)*RAND_MAX>rand())
            now_x=nxt_x,now_y=nxt_y;
        T*=0.998;
    }
}

int main() {
    srand(time(0));
    int T=read();
    while(T--) {
        lenx=read(),leny=read(),n=read();
        ans=-1e18;ansx=lenx/2;ansy=leny/2;
        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        Anneal();
        printf("The safest point is (%.1lf, %.1lf).\n",ansx,ansy);
    }
    return 0;
}

程序仅供参考

POJ1379 Run Away

标签：can   exp   代码   sqrt   safe   xxx   平面   卡死   i++   

原文地址：https://www.cnblogs.com/AKMer/p/9599381.html