uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1058

　　半平面交求面积最值。直接枚举C(20,8)的所有情况即可。

代码如下：

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #include <cmath>

 #include <vector>

 using namespace std;

 const int N = ;

 struct Point {

     double x, y;

     Point() {}

     Point(double x, double y) : x(x), y(y) {}

 } pt[N], poly[N];

 template<class T> T sqr(T x) { return x * x;}

 typedef Point Vec;

 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}

 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}

 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}

 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);}

 const double EPS = 1e-;

 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}

 inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}

 inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}

 inline double dotDet(Point o, Point a, Point b) { return dotDet(a - o, b - o);}

 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}

 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));}

 inline Vec normal(Vec x) { return Vec(-x.y, x.x) / vecLen(x);}

 struct DLine {

     Point p;

     Vec v;

     double ang;

     DLine() {}

     DLine(Point p, Vec v) : p(p), v(v) { ang = atan2(v.y, v.x);}

     bool operator < (const DLine &L) const { return ang < L.ang;}

     DLine move(double x) { return DLine(p + normal(v) * x, v);}

 } dl[N];

 inline bool onLeft(DLine L, Point p) { return crossDet(L.v, p - L.p) > ;}

 inline Point dLineIntersect(DLine a, DLine b) { return a.p + a.v * (crossDet(b.v, a.p - b.p) / crossDet(a.v, b.v));}

 struct Poly {

     vector<Point> pt;

     Poly() { pt.clear();}

     ~Poly() {}

     Poly(vector<Point> &pt) : pt(pt) {}

     Point operator [] (int x) const { return pt[x];}

     int size() { return pt.size();}

     double area() {

         if (pt.size() < ) return 0.0;

         double ret = 0.0;

         for (int i = , sz = pt.size(); i < sz; i++)

             ret += crossDet(pt[i], pt[(i + ) % sz]);

         return fabs(ret / 2.0);

     }

 } ;

 Point p[N];

 DLine q[N];

 int halfPlane(DLine *L, int n) {

     sort(L, L + n);

     int fi, la;

     q[fi = la = ] = L[];

     for (int i = ; i < n; i++) {

         while (fi < la && !onLeft(L[i], p[la - ])) la--;

         while (fi < la && !onLeft(L[i], p[fi])) fi++;

         q[++la] = L[i];

         if (fabs(crossDet(q[la].v, q[la - ].v)) < EPS) {

             la--;

             if (onLeft(q[la], L[i].p)) q[la] = L[i];

         }

         if (fi < la) p[la - ] = dLineIntersect(q[la - ], q[la]);

     }

     while (fi < la && !onLeft(q[fi], p[la - ])) la--;

     if (la < fi) return ;

     p[la] = dLineIntersect(q[la], q[fi]);

     int ret = ;

     for (int i = fi; i <= la; i++) poly[ret++] = p[i];

     return ret;

 }

 //Poly halfPlane(DLine *L, int n) {

 //    Poly ret = Poly();

 //    sort(L, L + n);

 //    int fi, la;

 //    Point *p = new Point[n];

 //    DLine *q = new DLine[n];

 //    q[fi = la = 0] = L[0];

 //    for (int i = 1; i < n; i++) {

 //        while (fi < la && !onLeft(L[i], p[la - 1])) la--;

 //        while (fi < la && !onLeft(L[i], p[fi])) fi++;

 //        q[++la] = L[i];

 //        if (fabs(crossDet(q[la].v, q[la - 1].v)) < EPS) {

 //            la--;

 //            if (onLeft(q[la], L[i].p)) q[la] = L[i];

 //        }

 //        if (fi < la) p[la - 1] = dLineIntersect(q[la - 1], q[la]);

 //    }

 //    while (fi < la && !onLeft(q[fi], p[la - 1])) la--;

 //    if (la < fi) return ret;

 //    p[la] = dLineIntersect(q[la], q[fi]);

 //    for (int i = fi; i <= la; i++) ret.pt.push_back(p[i]);

 //    return ret;

 //}

 double polyArea(Point *p, int n) {

     if (n < ) return 0.0;

     double sum = 0.0;

     p[n] = p[];

     Point O = Point(0.0, 0.0);

     for (int i = ; i < n; i++) sum += crossDet(O, p[i], p[i + ]);

     return fabs(sum / 2.0);

 }

 double work(int st, int n, double d) {

 //    cout << st << endl;

     for (int i = ; i < n; i++) dl[i] = DLine(pt[i], pt[i + ] - pt[i]).move((st & ( << i)) ==  ? 0.0 : d);

     int tmp = halfPlane(dl, n);

     return polyArea(poly, tmp);

 //    Poly tmp = halfPlane(dl, n);

 //    return tmp.area();

 }

 int cntBit(int x) {

     int cnt = ;

     while (x > ) {

         if (x & ) cnt++;

         x >>= ;

     }

     return cnt;

 }

 int main() {

 //    freopen("in", "r", stdin);

     int n, k;

     double d;

     while (~scanf("%d%d%lf", &n, &k, &d) && (n + k + d > EPS)) {

         for (int i = ; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y);

         pt[n] = pt[];

         double ans = 0.0, area = polyArea(pt, n);

         for (int i = , end =  << n; i < end; i++) {

             if (cntBit(i) <= k) {

 //                cout << i << endl;

                 ans = max(ans, area - work(i, n, d));

             }

             if (sgn(ans - area) == ) break;

         }

         printf("%.2f\n", ans);

     }

     return ;

 }

能通过POJ的代码：

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #include <cmath>

 #include <vector>

 using namespace std;

 const int N = ;

 struct Point {

     double x, y;

     Point() {}

     Point(double x, double y) : x(x), y(y) {}

 } pt[N], poly[N];

 template<class T> T sqr(T x) { return x * x;}

 typedef Point Vec;

 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}

 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}

 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}

 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);}

 const double EPS = 1e-;

 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}

 inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}

 inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}

 inline double dotDet(Point o, Point a, Point b) { return dotDet(a - o, b - o);}

 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}

 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));}

 inline Vec normal(Vec x) { return Vec(-x.y, x.x) / vecLen(x);}

 struct DLine {

     Point p;

     Vec v;

     double ang;

     DLine() {}

     DLine(Point p, Vec v) : p(p), v(v) { ang = atan2(v.y, v.x);}

     bool operator < (const DLine &L) const { return ang < L.ang;}

     DLine move(double x) { return DLine(p + normal(v) * x, v);}

 } dl[N];

 inline bool onLeft(DLine L, Point p) { return crossDet(L.v, p - L.p) > ;}

 inline Point dLineIntersect(DLine a, DLine b) { return a.p + a.v * (crossDet(b.v, a.p - b.p) / crossDet(a.v, b.v));}

 struct Poly {

     vector<Point> pt;

     Poly() { pt.clear();}

     ~Poly() {}

     Poly(vector<Point> &pt) : pt(pt) {}

     Point operator [] (int x) const { return pt[x];}

     int size() { return pt.size();}

     double area() {

         if (pt.size() < ) return 0.0;

         double ret = 0.0;

         for (int i = , sz = pt.size(); i < sz; i++)

             ret += crossDet(pt[i], pt[(i + ) % sz]);

         return fabs(ret / 2.0);

     }

 } ;

 Point p[N];

 DLine q[N], tmpDL[N];

 int halfPlane(DLine *L, int n) {

     for (int i = ; i < n; i++) tmpDL[i] = L[i];

     sort(L, L + n);

     int fi, la;

     q[fi = la = ] = L[];

     for (int i = ; i < n; i++) {

         while (fi < la && !onLeft(L[i], p[la - ])) la--;

         while (fi < la && !onLeft(L[i], p[fi])) fi++;

         q[++la] = L[i];

         if (fabs(crossDet(q[la].v, q[la - ].v)) < EPS) {

             la--;

             if (onLeft(q[la], L[i].p)) q[la] = L[i];

         }

         if (fi < la) p[la - ] = dLineIntersect(q[la - ], q[la]);

     }

     for (int i = ; i < n; i++) L[i] = tmpDL[i];

     while (fi < la && !onLeft(q[fi], p[la - ])) la--;

     if (la < fi) return ;

     p[la] = dLineIntersect(q[la], q[fi]);

     int ret = ;

     for (int i = fi; i <= la; i++) poly[ret++] = p[i];

     return ret;

 }

 double polyArea(Point *p, int n) {

     if (n < ) return 0.0;

     double sum = 0.0;

     p[n] = p[];

     Point O = Point(0.0, 0.0);

     for (int i = ; i < n; i++) sum += crossDet(O, p[i], p[i + ]);

     return fabs(sum / 2.0);

 }

 double work(int st, int n, double d) {

 //    cout << st << endl;

     for (int i = ; i < n; i++) dl[i] = DLine(pt[i], pt[i + ] - pt[i]).move((st & ( << i)) ==  ? 0.0 : d);

     int tmp = halfPlane(dl, n);

     return polyArea(poly, tmp);

 //    Poly tmp = halfPlane(dl, n);

 //    return tmp.area();

 }

 int cntBit(int x) {

     int cnt = ;

     while (x > ) {

         if (x & ) cnt++;

         x >>= ;

     }

     return cnt;

 }

 const double FINF = 1e100;

 double ans, d;

 void dfs(int id, int tt, int used, int k) {

     if (id >= tt) {

         int tmp = halfPlane(dl, tt);

         ans = min(ans, polyArea(poly, tmp));

         return ;

     }

     dl[id] = DLine(pt[id], pt[id + ] - pt[id]);

     dfs(id + , tt, used, k);

     if (used >= k) return ;

     dl[id] = DLine(pt[id], pt[id + ] - pt[id]).move(d);

     dfs(id + , tt, used + , k);

 }

 int main() {

 //    freopen("in", "r", stdin);

     int n, k;

     while (~scanf("%d%d%lf", &n, &k, &d) && (n + k + d > EPS)) {

         for (int i = ; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y);

         pt[n] = pt[];

         double area = polyArea(pt, n);

         ans = FINF;

         dfs(, n, , k);

         printf("%.2f\n", area - ans);

     }

     return ;

 }

——written by Lyon

poj 1271 && uva 10117 Nice Milk （半平面交）的更多相关文章

  1. poj 2451 Uyuw's Concert （半平面交）

    2451 -- Uyuw's Concert 继续半平面交,这还是简单的半平面交求面积,不过输入用cin超时了一次. 代码如下: #include <cstdio> #include &l ...

  2. poj 2451 Uyuw's Concert（半平面交）

    Uyuw's Concert Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 8580   Accepted: 3227 De ...

  3. POJ - 1474 ：Video Surveillance （半平面交-求核）

    pro:顺时针给定多边形,问是否可以放一个监控,可以监控到所有地方,即问是否存在多边形的核. 此题如果两点在同一边界上(且没有被隔段),也可以相互看到. sol:求多边形是否有核.先给直线按角度排序, ...

  4. POJ 3384 Feng Shui（计算几何の半平面交+最远点对）

    Description Feng shui is the ancient Chinese practice of placement and arrangement of space to achie ...

  5. POJ 3525/UVA 1396 Most Distant Point from the Sea（二分+半平面交）

    Description The main land of Japan called Honshu is an island surrounded by the sea. In such an isla ...

  6. poj 3335（半平面交）

    链接:http://poj.org/problem?id=3335     //大牛们常说的测模板题 ------------------------------------------------- ...

  7. poj 1755 半平面交+不等式

    Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6461   Accepted: 1643 Descrip ...

  8. poj 1279 半平面交核面积

    Art Gallery Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6668   Accepted: 2725 Descr ...

  9. poj 3335 Rotating Scoreboard（半平面交）

    Rotating Scoreboard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6420   Accepted: 25 ...

随机推荐

  1. 洛谷P1965 转圈游戏 [2013NOIP提高组 D1T1][2017年6月计划 数论04]

    P1965 转圈游戏 题目描述 n 个小伙伴(编号从 0 到 n-1)围坐一圈玩游戏.按照顺时针方向给 n 个位置编号,从0 到 n-1.最初,第 0 号小伙伴在第 0 号位置,第 1 号小伙伴在第 ...

  2. 微信小程序之组件开发中的基础知识

    跟着视频开始小程序的项目的开发,视频中这个小程序已经上线了,可以很好的看着小程序的界面进行开发,昨天看了一下具体的需求,觉得真的细节好多啊,而且其中设计的组件的思想也是很好的,能够很好的实现代码的复用 ...

  3. Django项目：CRM(客户关系管理系统)--15--07PerfectCRM实现King_admin显示注册的表01

    <th ><a href="/kingadmin/{% get_app_name admin_class.model %}/{% get_model_name admin_ ...

  4. bzoj 1024 [SCOI2009]生日快乐——模拟

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=1024 可以枚举这边放多少块.那边放多少块. 注意精度.不要每次用x*y/base算有多少块, ...

  5. 批处理启动应用程序（win）

    @echo off net session >nul 2>&1 " ( echo Oops: This tools must run with administrator ...

  6. 微信小程序--轮播图，标题，盒子，tab栏的合成例子

    小程序是什么? 微信小程序,是一种不需要下载安装即可使用的应用,用户扫一扫或搜一下即可打开应用,在微信-发现-小程序可打开应用. 一.小程序的样式编写: 目录结构: app.json { " ...

  7. 提交方式get和post有什么区别

    提交方式post和get有什么区别? (1)post是向服务器传送数据:get是从服务器上获取数据. (2)在客户端,get是把参数数据队列加到提交表单的ACTION属性所指的URL中,值和表单内各个 ...

  8. AC自动机fail树小结

    建议大家学过AC自动机之后再来看这篇小结 fail树就是讲fail指针看做一条边连成的树形结构 fail指针在AC自动机中的含义是指以x为结尾的后缀在其他模式串中所能匹配的最长前缀的长度 所以在模式串 ...

  9. Linux之Shell1

    1.输出命令:echo echo [选项] [输出内容] : -e  支持反斜线控制的字符转换.(类似于C语言的\) \\ 输出\本身 \t Tab键 \n 换行符 \f 换页符 ...  

  10. 一条SQL完成跨数据库实例Join查询

    背景 随着业务复杂程度的提高.数据规模的增长,越来越多的公司选择对其在线业务数据库进行垂直或水平拆分,甚至选择不同的数据库类型以满足其业务需求.原本在同一数据库实例里就能实现的SQL查询,现在需要跨多 ...