Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 41


Problem Description

Give you a simple polygon and a line, and your task is to calculate the length of the line which is covered by the polygon.

 

Input

There are several test cases.
Each test case starts with one non-negative integer n (3 <= n <= 1000) giving the number of the vertices of the polygon. Then following n lines, each line contains two real numbers standing for the x and y coordinates of a vertex. The vertices are given either in clockwise or counterclockwise order. Then four real numbers (sx, sy, tx, ty), standing for the coordinates of the two points of the line. 
All real numbers are between -100000 and 100000.

 

Output

For each case print the total length of the line that covered by the polygon, with 3 digits after the decimal point.

 

Sample Input


4 0 0 0 1 1 1 1 0 0 0 1 1 4 0 0 0 1 1 1 1 0 0 0 1 0 9 0 0 0 2 1 1 2 2 3 1 4 2 5 1 6 2 6 0 0 1 6 1


Sample Output

1.414 1.000 6.000

 


总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。

​


 


#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct Point
{
    double x;
    double y;
} p[1001], px[10001];
int n;
double eps=1e-8;
int cmp(Point a, Point b)
{
    if(abs(a.x-b.x)<eps)
        return a.y<b.y;
    return a.x<b.x;
}
double dist(Point a,Point b)
{
    return sqrt((a.x - b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double direction(Point pi,Point pj,Point pk)
{
    return (pk.x-pi.x)*(pj.y-pi.y)-(pj.x-pi.x)*(pk.y-pi.y);
}
bool on_segment(Point pi,Point pj,Point pk)
{
    if(direction(pi, pj, pk)==0)
    {
        if(pk.x>=min(pi.x,pj.x)&&pk.x<=max(pi.x,pj.x)&&pk.y>=min(pi.y,pj.y)&&pk.y<=max(pi.y,pj.y))
            return true;
    }
    return false;
}
bool segments_intersect(Point p1,Point p2,Point p3,Point p4)
{
    double d1=direction(p3,p4,p1);
    double d2=direction(p3,p4,p2);
    double d3=direction(p1,p2,p3);
    double d4=direction(p1,p2,p4);
    if(d1*d2<0&&d3*d4<0)
        return true;
    else if(d1==0&&on_segment(p3,p4,p1))
        return true;
    else if(d2==0&&on_segment(p3,p4,p2))
        return true;
    else if(d3==0&&on_segment(p1,p2,p3))
        return true;
    else if(d4==0&&on_segment(p1,p2,p4))
        return true;
    return false;
}
Point intersection(Point a1, Point a2, Point b1, Point b2)
{
    Point ret = a1;
    double t = ((a1.x - b1.x) * (b1.y - b2.y) - (a1.y - b1.y) * (b1.x - b2.x))
               / ((a1.x - a2.x) * (b1.y - b2.y) - (a1.y - a2.y) * (b1.x - b2.x));
    ret.x += (a2.x - a1.x) * t;
    ret.y += (a2.y - a1.y) * t;
    return ret;
}
int InPolygon(Point a)
{
    int i;
    Point b,c,d;
    b.y=a.y;
    b.x=1e15;
    int count=0;
    for(i=0; i<n; i++)
    {
        c = p[i];
        d = p[i + 1];
        if(on_segment(c,d,a))
            return 1;
        if(abs(c.y-d.y)<eps)
            continue;
        if(on_segment(a,b,c))
        {
            if(c.y>d.y)
                count++;
        }
        else if(on_segment(a,b,d))
        {
            if(d.y>c.y)
                count++;
        }
        else if(segments_intersect(a,b,c,d))
            count++;
    }
    return count%2;
}
bool Intersect(Point s,Point e,Point a,Point b)
{
    return direction(e,a,s)*direction(e,b,s)<=0;
}
double calculate(Point s,Point e)
{
    int i,k=0;
    double sum;
    Point a,b,temp;
    for(i=0; i<n; i++)
    {
        a=p[i];
        b=p[i+1];
        if(abs(direction(e,a,s))<eps&&abs(direction(e,b,s))<eps)
        {
            px[k++]=a;
            px[k++]=b;
        }
        else if(Intersect(s,e,a,b))
        {
            px[k++]=intersection(s,e,a,b);
        }
    }
    if(k==0)
        return 0.0;
    sort(px,px+k,cmp);
    px[k]=px[0];
    sum=0;
    for(i=0; i<k-1; i++)
    {
        a=px[i];
        b=px[i+1];
        temp.x=(a.x+b.x)/2.0;
        temp.y=(a.y+b.y)/2.0;
        if(InPolygon(temp))
            sum+=dist(a,b);
    }
    return sum;
}
int main()
{
    int i;
    double sum;
    Point s,e;
    while(~scanf("%d",&n)&&n)
    {
        for(i=0; i<n; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        p[n]=p[0];
        scanf("%lf%lf%lf%lf",&s.x,&s.y,&e.x,&e.y);
        sum=calculate(s,e);
        printf("%.3lf\n",sum);
    }
    return 0;
}

@执念  "@☆但求“❤”安★ 下次我们做的一定会更好。。。。


为什么这次的题目是英文的。。。。QAQ...