UVA 10060 A Hole to Catch a Man(求多边形面积相关)

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http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=12&page=show_problem&problem=1001

题意:

给你n个厚度已知的多边形钢板,现在要你融化它们来做成尽量多的(厚度已知)圆形井盖. 问你最多能做成几个?

分析:

本题比较水.

直接求出每个多边形钢板的体积,然后算出体积和sum. 之后输出sum/每个圆形井盖的体积就是所求.

求多边形的面积用刘汝佳的模板即可.

AC代码:

#include<cstdio>#include<cmath>using namespace std;const double PI=acos(-1.0);const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps) return 0;    return x<0?-1:1;}struct Point{    double x,y;    Point(){}    Point(double x,double y):x(x),y(y){}}P[1010];typedef Point Vector;Vector operator-(Point A,Point B){    return Vector(A.x-B.x, A.y-B.y);}double Cross(Vector A,Vector B){    return A.x*B.y-A.y*B.x;}double PolygenArea(Point *p,int n){    double area=0;    for(int i=1;i<n-1;++i)        area += Cross(p[i]-p[0],p[i+1]-p[0]);    return fabs(area)/2;}int main(){    int num;    while(scanf("%d",&num)==1 && num)    {        double v=0;//总体积        while(num--)        {            double thickness;//厚度            int n=0;//当前多边形点数            scanf("%lf",&thickness);            scanf("%lf%lf",&P[0].x,&P[0].y);            while(++n)            {                scanf("%lf%lf",&P[n].x,&P[n].y);                if(dcmp(P[n].x-P[0].x)==0 && dcmp(P[n].y-P[0].y)==0)  break;                //上面写成了dcmp(P[n].x-P[0].y)==0 WA了我半天 我擦            }            v += PolygenArea(P,n)*thickness;        }        double r,thick;//圆的半径和厚度        scanf("%lf%lf",&r,&thick);        printf("%d\n", (int)( v/(PI*r*r*thick) ) );    }    return 0;}

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