模拟退火合集

hdu 3644 A Chocolate Manufacturer's Problem

首次接触模拟退火，看来还是挺神奇的。

主要参考这篇博文：http://blog.sina.com.cn/s/blog_7da04dd30100wlqj.html

题意判断多边形内部能否容纳一个半径为R的圆，即在有限的平面内找最优范围。遗传算法的结果难以掌控，爬山算法又没法保证跳出局部最优，所以基于贪心原则的模拟退火算法还是值得考虑的。

然后就是设定每次变化的步长和演化方式。该题可以从每条边的中点开始，向一个随机方向延伸，使找出的点到各条边的最小距离最大。

#include <stdio.h>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
#include <set>
#include <iostream>
#include <cmath>
using namespace std;
#ifdef __GNUC__
typedef long long LL;
inline void opt64(LL a) {
    printf("%lld", a);
}
#else
typedef __int64 LL;
inline void opt64(LL a) {
    printf("%I64d", a);
}
#endif
const int MAXN = 55, maxstps = 6;
const double inf = 1e20, eps = 1e-3, aeps = 1e-3;
int n;
double radius;
inline int cmp(double x, double y) {
    return ((x-y)>aeps) - ((x-y)<-aeps);
}
inline int cmp(double x) {
    return (x>aeps) - (x<-aeps);
}
struct _point
{
    double x, y, r;
    _point(double xx=0.0, double yy=0.0):x(xx),y(yy) {}
    void ipt() {
        scanf("%lf%lf", &x, &y);
    }
    double dot(const _point &a) {
        return x*a.x+y*a.y;
    }
    double cross(const _point &a) {
        return x*a.y-y*a.x;
    }
    _point operator + (const _point &a) {
        return _point(x+a.x, y+a.y);
    }
    _point operator - (const _point &a) {
        return _point(x-a.x, y-a.y);
    }
    double dis() {
        return sqrt(x*x + y*y);
    }
} da[MAXN], db[MAXN];
bool point_in_poly(_point p)
{
    _point q = p, a, b;
    q.x = inf;
    int res = 0;
    double x;
    for (int i = 0; i< n; ++i)
    {
        a = da[i];
        b = da[1+i];
        if (cmp(a.y, b.y) == 0 ||
            cmp(p.y, min(a.y, b.y)) == -1 ||
            cmp(p.y, max(a.y, b.y)) >= 0 ) continue;
        x = (p.y-a.y)*(b.x - a.x) / (b.y - a.y) + a.x;
        if (cmp(x, p.x) == 0) return 0;
        if (cmp(x, p.x) == 1) ++res;
    }
    return res%2;
}
double dis_Point_to_seg(_point o, _point a, _point b)
{
    _point l1 = o-a, l2 = o-b, l3 = a-b;
    if (cmp(l1.dot(l3)) == 1) return l1.dis();
    if (cmp(l2.dot(l3)) == -1) return l2.dis();
    return fabs(l1.cross(l2)) / l3.dis();
}
void cal(_point * a)
{
    a->r = inf;
    for (int i = 0; i< n; ++i)
        a->r = min( dis_Point_to_seg(*a, da[i], da[i+1]), a->r);
}
int solve(double r)
{
    _point tmp;
    double agl;
    while (r > eps)
    {
        for (int i = 0; i< n; ++i)
        {
            for (int j = 0; j< maxstps; ++j)
            {
                agl = rand();
                tmp.x = db[i].x + cos(agl)*r;
                tmp.y = db[i].y + sin(agl)*r;
                if (point_in_poly(tmp))
                {
                    cal(&tmp);
                    if (tmp.r > db[i].r)
                    {
                        db[i] = tmp;
                        if (cmp(tmp.r, radius) >= 0)
                            return 1;
                    }
                }

            }
        }
        r *= 0.9;
    }
    return 0;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    while (scanf("%d", &n) && n)
    {
        double maxx = -1, maxy = -1, minx = inf, miny = inf;
        for (int i = 0; i< n; ++i)
            da[i].ipt(), maxx = max(maxx, da[i].x), maxy = max(maxy, da[i].y),
                         minx = min(minx, da[i].x), miny = min(miny, da[i].y);
        da[n] = da[0];
        for (int i = 0; i< n; ++i)
        {
            db[i].x = (da[i].x + da[i+1].x)/2.0;
            db[i].y = (da[i].y + da[i+1].y)/2.0;
            db[i].r = 0.0;
        }
        maxx = maxx - minx;
        maxy = maxy - miny;
        scanf("%lf", &radius);
        if (solve(sqrt(maxx*maxx + maxy*maxy)/2.0)) puts("Yes");
        else puts("No");
    }
    return 0;
}

hdu 5017 Ellipsoid

找出椭球面上距离球心最近的点

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const double eps = 1e-10, r = 0.99;
double a, b, c, d, e, f, res;
double dir[8][2] = {0,1, 0,-1, -1,0, 1,0, -1,1, -1,-1, 1,1, 1,-1};
inline double dis (double x, double y, double z)
{
    return sqrt(x*x + y*y + z*z);
}
inline int cmp(double x)
{
    return (x>eps) - (x<-eps);
}
int cal(double A, double B, double C, double &z)
{
    double q = B*B - 4*A*C;
    if (cmp(q) < 0) return 0;
    z = (sqrt(q) - B)/A/2.0;
    return 1;
}
void solve()
{
    double x=0, y=0, z, k = 1.0, ax=0, ay=0, pp;
    cal(c, 0, -1, z);
    res = dis(0, 0, z);
    while (k > eps)
    {
        x = ax; y = ay;
        for (int i = 0; i< 8; ++i)
        {
            double xx = x+dir[i][0]*k, yy = y+dir[i][1]*k;
            if (cal(c, d*yy+e*xx, a*xx*xx+b*yy*yy+f*xx*yy-1, z) == 0) continue;
            if ( cmp((pp=dis(xx, yy, z)) - res) < 0)
                ax = xx, ay = yy, res = pp;
        }
        k *= r;
    }
    printf("%.7lf\n", res);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    while (scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &e, &f) != EOF)
    {
        solve();
    }
    return 0;
}

hdu 3932 Groundhog Build Home

最小覆盖圆

#include <stdio.h>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
#include <set>
#include <iostream>
#include <cmath>
using namespace std;
#ifdef __GNUC__
typedef long long LL;
inline void opt64(LL a) {
    printf("%lld", a);
}
#else
typedef __int64 LL;
inline void opt64(LL a) {
    printf("%I64d", a);
}
#endif
const int MAXN = 1005, maxstps = 5;
const double inf = 1e20, eps = 1e-4, aeps = 1e-6;
int n;
double radius;
double maxx = -1, maxy = -1, minx = inf, miny = inf;
inline int cmp(double x, double y) {
    return ((x-y)>aeps) - ((x-y)<-aeps);
}
inline int cmp(double x) {
    return (x>aeps) - (x<-aeps);
}
struct _point
{
    double x, y;
    _point(double xx=0.0, double yy=0.0):x(xx),y(yy) {}
    void ipt() {
        scanf("%lf%lf", &x, &y);
    }
    _point operator - (const _point &a) {
        return _point(x-a.x, y-a.y);
    }
    double dis() {
        return sqrt(x*x + y*y);
    }
} da[MAXN];
inline bool point_in_rect(double x, double y)
{
    if (cmp(x-maxx) == 1 ||
        cmp(x-minx) == -1 ||
        cmp(y-maxy) == 1 ||
        cmp(y-miny) == -1) return 0;
    return 1;
}
void solve(double x, double y, double r, double mxdis)
{
    double angle, xx=x, yy=y, ax, ay, d, c;
    while (r > eps)
    {
        x = xx; y = yy;
        for (angle = 0.0; angle< 6.28; angle += 0.1)
        {
            ax = x+cos(angle)*r; ay=y+sin(angle)*r;
            if (point_in_rect(ax,ay))
            {
                _point ap(ax,ay);
                c = -1;
                for (int j = 0; j< n; ++j)
                {
                    d = (ap-da[j]).dis();
                    if (cmp(d-c)==1)
                        c = d;
                }
                if (cmp(c-mxdis)==-1)
                {
                    mxdis = c;
                    xx = ax; yy = ay;
                }
            }
        }
        r *= 0.7;
    }
    printf("(%.1lf,%.1lf).\n", x, y);
    printf("%.1lf\n", mxdis);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    while (scanf("%lf%lf%d", &maxx, &maxy, &n) != EOF)
    {
        minx = 0.0, miny = 0.0;
        double x1=inf, x2=-1, y1=inf, y2=-1, c=-1;
        for (int i = 0; i< n; ++i)
            da[i].ipt(), x1=min(x1,da[i].x),
            x2=max(x2, da[i].x), y1=min(y1,da[i].y), y2=max(y2,da[i].y);
        _point pp((x1+x2)/2, (y1+y2)/2);
        for (int i = 0; i< n; ++i)
            c = max(c, (pp-da[i]).dis());
        solve(pp.x, pp.y, sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))*0.5, c);
    }
    return 0;
}