poj1039——计算几何 求直线与线段交点，判断两条直线是否相交

poj1039——计算几何  求直线与线段交点，判断两条直线是否相交

Pipe

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9439		Accepted: 2854

Description

The GX Light Pipeline Company started to prepare bent pipes for the new transgalactic light pipeline. During the design phase of the new pipe shape the company ran into the problem of determining how far the light can reach inside each component of the pipe. Note that the material which the pipe is made from is not transparent and not light reflecting. 

Each pipe component consists of many straight pipes connected tightly together. For the programming purposes, the company developed the description of each component as a sequence of points [x1; y1], [x2; y2], . . ., [xn; yn], where x1 < x2 < . . . xn . These are the upper points of the pipe contour. The bottom points of the pipe contour consist of points with y-coordinate decreased by 1. To each upper point [xi; yi] there is a corresponding bottom point [xi; (yi)-1] (see picture above). The company wants to find, for each pipe component, the point with maximal x-coordinate that the light will reach. The light is emitted by a segment source with endpoints [x1; (y1)-1] and [x1; y1] (endpoints are emitting light too). Assume that the light is not bent at the pipe bent points and the bent points do not stop the light beam.

Input

The input file contains several blocks each describing one pipe component. Each block starts with the number of bent points 2 <= n <= 20 on separate line. Each of the next n lines contains a pair of real values xi, yi separated by space. The last block is denoted with n = 0.

Output

The output file contains lines corresponding to blocks in input file. To each block in the input file there is one line in the output file. Each such line contains either a real value, written with precision of two decimal places, or the message Through all the pipe.. The real value is the desired maximal x-coordinate of the point where the light can reach from the source for corresponding pipe component. If this value equals to xn, then the message Through all the pipe. will appear in the output file.

Sample Input

4
0 1
2 2
4 1
6 4
6
0 1
2 -0.6
5 -4.45
7 -5.57
12 -10.8
17 -16.55
0

Sample Output

4.67
Through all the pipe.

题意：求光线通过管道的最远坐标
思路：能通过最远的一条光线一定经过一个up[i]和一个down[j]，枚举即可。通过判断直线up[i]down[j]是否与第k个管口线段up[k]down[k]相交来判断是否能通过管口k，从小到大遍历，当碰到不能通过时，即能通过第k-1个而不能通过第k个，最远坐标一定在xk和xk-1之间，光线与k-1和k之间的上下管壁交点横坐标的最大值。
判断直线与线段相交用叉积来判断左右拐从而判断相交，求直线与线段交点则用定比分点，其中线段长度比值用叉积求面积比代替
hack点：求直线与线段交点时需先排除直线与线段不相交的情况

#include
#include
#include
#include
#include
#include

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);
const double eps=0.00000001;

int n;
struct Point
{
    double x,y;
    friend double operator*(const Point A,const Point B)
    {
        return A.x*B.y-A.y*B.x;
    }
    friend Point operator-(const Point A,const Point B)
    {
        return {A.x-B.x,A.y-B.y};
    }
};
Point up[maxn],down[maxn];

double MAX(double a,double b)
{
    return a>b+eps?a:b;
}

bool intersect(Point A,Point B,Point C,Point D)///判断直线AB是否与线段CD相交
{
    if(((C-A)*(B-A))*((D-A)*(B-A))return true;
    return false;
}

Point inter_point(Point A,Point B,Point C,Point D)///返回直线AB和线段CD的交点
{
    if(!intersect(A,B,C,D)) return {-INF*1.0,0};///判断直线AB是否与线段CD相交，不相交必须须特判
    double area1=fabs((B-A)*(C-A));
    double area2=fabs((B-A)*(D-A));
    double x=(area1*D.x+area2*C.x)/(area1+area2);
    double y=(area1*D.y+area2*C.y)/(area1+area2);
    return {x,y};
}

int main()
{
    while(cin>>n,n){
        for(int i=0;i){
            cin>>up[i].x>>up[i].y;
            down[i]={up[i].x,up[i].y-1.0};
        }
        double ans=-INF*1.0;
        bool flag=0;
        for(int i=0;i){
            for(int j=0;j){
                if(i==j) continue;
                if(!intersect(up[i],down[j],up[0],down[0])) continue;
                for(int k=1;k){
                    if(!intersect(up[i],down[j],up[k],down[k])){
                        Point A=inter_point(up[i],down[j],up[k-1],up[k]);
                        Point B=inter_point(up[i],down[j],down[k-1],down[k]);
                        double Max=MAX(A.x,B.x);
                        if(Max>ans+eps) ans=Max;
                        //cout<<"i="<
                        break;
                    }
                    else if(k==n-1){
                        flag=1;break;
                    }
                }
                if(flag) break;
            }
            if(flag) break;
        }
        if(flag) printf("Through all the pipe.\n");
        else printf("%.2f\n",ans);
    }
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/--560/p/4387755.html