poj1113Wall(凸包）

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顺便整理出来一份自己看着比较顺眼的模板

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 1010
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = 3.141592653;
 16 const double inf = ~0u>>2;
 17 struct Point
 18 {
 19     double x,y;
 20     Point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写
 21 }p[N],ch[N];
 22 typedef Point pointt;
 23 pointt operator + (Point a,Point b)
 24 {
 25     return Point(a.x+b.x,a.y+b.y);
 26 }
 27 pointt operator - (Point a,Point b)
 28 {
 29     return Point(a.x-b.x,a.y-b.y);
 30 }
 31 pointt operator * (Point a,double b)
 32 {
 33     return Point(a.x*b,a.y*b);
 34 }
 35 pointt operator / (Point a,double b)
 36 {
 37     return Point(a.x/b,a.y/b);
 38 }
 39 bool operator < (const Point &a,const Point &b)
 40 {
 41     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 42 }
 43 int dcmp(double x)
 44 {
 45     if(fabs(x)<eps) return 0;
 46     else return x<0?-1:1;
 47 }
 48 bool operator == (const Point &a,const Point &b)
 49 {
 50     return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
 51 }
 52 //求点积以及利用点积求长度和夹角的函数
 53 double dot(Point a,Point b)
 54 {
 55     return a.x*b.x+a.y*b.y;
 56 }
 57 //叉积及叉积求面积
 58 double cross(Point a,Point b)
 59 {
 60     return a.x*b.y-a.y*b.x;
 61 }
 62 double  dis(Point a)
 63 {
 64     return sqrt(dot(a,a));
 65 }
 66 double mul(Point p0,Point p1,Point p2)
 67 {
 68     return cross(p1-p0,p2-p0);
 69 }
 70 bool cmp(Point a,Point b)
 71 {
 72     if(dcmp(mul(p[1],a,b))==0)
 73         return dis(a-p[1])<dis(b-p[1]);
 74     else
 75         return dcmp(mul(p[1],a,b))>0;
 76 }
 77 int Graham(int n)
 78 {
 79     int i,k = 1,top = 1;
 80     Point tmp;
 81     if(n<=3)
 82     {
 83         for(i = 1; i <= 3 ; i++)
 84             ch[i] = p[i];
 85         return n+1;
 86     }
 87     for(i = 1 ; i <= n; i++)
 88     {
 89         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))
 90             k = i;
 91     }
 92     if(k!=1)
 93     {
 94         tmp = p[1];  p[1] = p[k];  p[k] = tmp;
 95     }
 96     sort(p+2,p+n+1,cmp);
 97     ch[top++] = p[1];
 98     ch[top++] = p[2];
 99     ch[top++] = p[3];
100     for(i = 4; i <= n ;)
101     {
102         if(top<3||mul(ch[top-2],ch[top-1],p[i])>0)
103         {
104             ch[top++] = p[i++];
105         }
106         else  top--;
107     }
108     return top;
109 }
110 int main()
111 {
112     int i,n,r;
113     while(scanf("%d%d",&n,&r)!=EOF)
114     {
115         for(i = 1 ; i <= n ; i++)
116         {
117             scanf("%lf%lf",&p[i].x,&p[i].y);
118         }
119         int top = Graham(n);
120         ch[top] = p[1];
121         double ma;
122         ma = 2*pi*r;
123         for(i = 1 ; i < top ; i++)
124         {
125             ma += dis(ch[i]-ch[i+1]);
126         }
127         printf("%.0f\n",ma);
128     }
129     return 0;
130 }
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posted @ 2014-06-25 15:29  _雨  阅读(142)  评论(0)  编辑  收藏  举报