hdu 1392 Surround the Trees
2015-04-01 10:06
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本题是一个关于几何的题目,题目的意思很简单,就是求最外围的长度,主要是 凸包的模版的运用,借鉴大神们的思路和模版,还有叉积的知识,然后a~~
#include <stdio.h> #include <math.h> #include <algorithm> #define Max 501 #define eps 1e-8 using namespace std; struct point { double x; double y; double jiao; }; point dingdian[Max]; //存储各个顶点的坐标 point points[Max]; int n; int size; double sum; int com1(point a,point b) //排序yx的点 的坐标 { if(a.y==b.y) { return a.x<b.x; } return a.y<b.y; } int com2(point a,point b) //排序角度小到大 { if(a.jiao==b.jiao) return a.x<b.x; return a.jiao<b.jiao; } void jijiao() //求与基点的角度 { int i; double x,y; for(i=1;i<n;i++) { x=points[i].x-points[0].x; y=points[i].y-points[0].y; points[i].jiao=acos((x/(sqrt(x*x+y*y)))); } } bool chaji(point A,point B,point C) //由判断是否向右拐 { double x1,x2,y1,y2; double jieguo; x1=B.x-A.x; x2=C.x-A.x; y1=B.y-A.y; y2=C.y-A.y; jieguo=x1*y2-x2*y1; if(fabs(jieguo)<eps) jieguo=0; if(jieguo<=0) return true; return false; } void zhaodingdian() //遍历寻找外围的顶点 并存储 { int top=0,i; sort(points,points+n,com1); jijiao(); sort(points+1,points+n,com2); dingdian[0]=points[0]; size=1; for(i=1;i<n;i++) { while(size>1&&chaji(dingdian[top-1],dingdian[top],points[i])) //循环取舍点 { size--; top--; } dingdian[++top]=points[i]; size++; } } double jisuan(point a,point b) //求距离 { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } int main() { int i; while(~scanf("%d",&n)&&n) { double sum=0; for(i=0;i<n;i++) { scanf("%lf%lf",&points[i].x,&points[i].y); } zhaodingdian(); for(i=0;i<size-1;i++) { sum+=jisuan(dingdian[i],dingdian[i+1]); } sum+=jisuan(dingdian[0],dingdian[size-1]); printf("%.2lf\n",sum); } return 0; }
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