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  • uva 10652

    大意:有n块矩形木板,你的任务是用一个面积尽量小的凸多边形把它们包起来,并计算出木板站整个包装面积的百分比。

    思路:按照题意将所有矩形顶点坐标存起来,旋转时先旋转从中心出发的向量,求得各个坐标之后,求凸包即可。

    水。。。。

      1 #include<cstdio>
      2 #include<cmath>
      3 #include<cstring>
      4 #include<algorithm>
      5 #include<iostream>
      6 #include<memory.h>
      7 #include<cstdlib>
      8 #include<vector>
      9 #define clc(a,b) memset(a,b,sizeof(a))
     10 #define LL long long int
     11 #define up(i,x,y) for(i=x;i<=y;i++)
     12 #define w(a) while(a)
     13 using namespace std;
     14 const int inf=0x3f3f3f3f;
     15 const int N = 4010;
     16 const double eps = 5*1e-13;
     17 const double pi = acos(-1);
     18 
     19 const double PI = acos(-1.0);
     20 double torad(double deg)
     21 {
     22     return deg/180 * PI;
     23 }
     24 
     25 struct Point
     26 {
     27     double x, y;
     28     Point(double x=0, double y=0):x(x),y(y) { }
     29 };
     30 
     31 typedef Point Vector;
     32 
     33 Vector operator + (const Vector& A, const Vector& B)
     34 {
     35     return Vector(A.x+B.x, A.y+B.y);
     36 }
     37 Vector operator - (const Point& A, const Point& B)
     38 {
     39     return Vector(A.x-B.x, A.y-B.y);
     40 }
     41 double Cross(const Vector& A, const Vector& B)
     42 {
     43     return A.x*B.y - A.y*B.x;
     44 }
     45 
     46 Vector Rotate(const Vector& A, double rad)
     47 {
     48     return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
     49 }
     50 
     51 bool operator < (const Point& p1, const Point& p2)
     52 {
     53     return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
     54 }
     55 
     56 bool operator == (const Point& p1, const Point& p2)
     57 {
     58     return p1.x == p2.x && p1.y == p2.y;
     59 }
     60 
     61 // 点集凸包
     62 // 如果不希望在凸包的边上有输入点,把两个 <= 改成 <
     63 // 如果不介意点集被修改,可以改成传递引用
     64 vector<Point> ConvexHull(vector<Point> p)
     65 {
     66     // 预处理,删除重复点
     67     sort(p.begin(), p.end());
     68     p.erase(unique(p.begin(), p.end()), p.end());
     69 
     70     int n = p.size();
     71     int m = 0;
     72     vector<Point> ch(n+1);
     73     for(int i = 0; i < n; i++)
     74     {
     75         while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
     76         ch[m++] = p[i];
     77     }
     78     int k = m;
     79     for(int i = n-2; i >= 0; i--)
     80     {
     81         while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
     82         ch[m++] = p[i];
     83     }
     84     if(n > 1) m--;
     85     ch.resize(m);
     86     return ch;
     87 }
     88 
     89 // 多边形的有向面积
     90 double PolygonArea(vector<Point> p)
     91 {
     92     double area = 0;
     93     int n = p.size();
     94     for(int i = 1; i < n-1; i++)
     95         area += Cross(p[i]-p[0], p[i+1]-p[0]);
     96     return area/2;
     97 }
     98 
     99 int main()
    100 {
    101     int T;
    102     scanf("%d", &T);
    103     while(T--)
    104     {
    105         int n;
    106         double area1 = 0;
    107         scanf("%d", &n);
    108         vector<Point> P;
    109         for(int i = 0; i < n; i++)
    110         {
    111             double x, y, w, h, j, ang;
    112             scanf("%lf%lf%lf%lf%lf", &x, &y, &w, &h, &j);
    113             Point o(x,y);
    114             ang = -torad(j);
    115             P.push_back(o + Rotate(Vector(-w/2,-h/2), ang));
    116             P.push_back(o + Rotate(Vector(w/2,-h/2), ang));
    117             P.push_back(o + Rotate(Vector(-w/2,h/2), ang));
    118             P.push_back(o + Rotate(Vector(w/2,h/2), ang));
    119             area1 += w*h;
    120         }
    121         double area2 = PolygonArea(ConvexHull(P));
    122         printf("%.1lf %%
    ", area1*100/area2);
    123     }
    124     return 0;
    125 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ITUPC/p/4881105.html
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