Problem D: Cutting tabletops

Bever
Lumber hires beavers to cut wood. The company has recently received a shippment of tabletops. Each tabletop is a convex polygon. However, in this hard economic times of cutting costs the company has ordered
the tabletops from a not very respectable but cheap supplier. Some of the tabletops have the right shape but they are slightly too big. The beavers have to chomp of a strip of wood of a fixed width from each edge of the tabletop such that they get a tabletop
of a similar shape but smaller. Your task is to find the area of the tabletop after beavers are done.

Input consists of a number of cases each presented on a separate line. Each line consists of a sequence of numbers. The first number is d the width of the strip of wood to be cut off of each edge
of the tabletop in centimeters. The next number n is an integer giving the number of vertices of the polygon. The next npairs of numbers present x_i and y_i coordinates
of polygon vertices for 1 <= i <= n given in clockwise order. A line containing only two zeroes terminate the input.

d is much smaller than any of the sides of the polygon. The beavers cut the edges one after another and after each cut the number of vertices of the tabletop is the same.

For each line of input produce one line of output containing one number to three decimal digits in the fraction giving the area of the tabletop after cutting.

Sample input

2 4 0 0 0 5 5 5 5 0

1 3 0 0 0 5 5 0

1 3 0 0 3 5.1961524 6 0

3 4 0 -10 -10 0 0 10 10 0

0 0

Output for sample input

1.000

1.257

2.785

66.294

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Problem Setter: Piotr Rudnicki

题目大意：

顺时针给定你一个凸多边形。问你削去d距离后，这个凸多边形的面积

解题思路：

也就是原来的凸包面积减去全部以凸包边为长度高为d的矩形面积加上多去除的部分（也就是1，2，3，4，5的面积），就是答案

解题代码：

#include <iostream>

#include <cstdio>

#include <vector>

#include <cmath>

#include <algorithm>

using namespace std;

struct point{

    double x,y;

    point(double x0=0,double y0=0){x=x0;y=y0;}

    double xchen(point p){//this X P

        return x*p.y-p.x*y;

    }

    double dchen(point p){//this X P

        return x*p.x+y*p.y;

    }

    double getlen(){

        return sqrt ( x*x+y*y );

    }

    double getdis(point p){

        return sqrt( (x-p.x)*(x-p.x) + (y-p.y)*(y-p.y) );

    }

};

const double eps=1e-7;

double d;

int n;

vector <point> p;

void input(){

    p.resize(n);

    for(int i=0;i<n;i++){

        scanf("%lf%lf",&p[i].x,&p[i].y);

    }

}

void solve(){

    double sum=0;

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

        point p1=point(p[i].x-p[0].x,p[i].y-p[0].y);

        point p2=point(p[i+1].x-p[0].x,p[i+1].y-p[0].y);

        sum+=fabs(p2.xchen(p1))/2.0;

    }

    for(int i=0;i<n;i++){

        double dis=p[i].getdis(p[(i+n-1)%n]);

        sum-=dis*d;

    }

    for(int i=0;i<n;i++){

        int t1=((i-1)+n)%n,t2=((i+1)+n)%n;

        point p1=point(p[t1].x-p[i].x,p[t1].y-p[i].y);

        point p2=point(p[t2].x-p[i].x,p[t2].y-p[i].y);

        double degree=acos( p1.dchen(p2)/p1.getlen()/p2.getlen() ) /2.0;

        double area=d/(tan(degree) )*d;

        sum+=area;

    }

    printf("%.3lf\n",sum);

}

int main(){

    while(scanf("%lf%d",&d,&n)!=EOF){

        if(fabs(d-0.0)<eps && n==0) break;

        input();

        solve();

    }

    return 0;

}