【解题报告】BUPT Online Judge Volume 9 Problem 1805 Segments【计算几何】（水题。因为不会用叉积做，自玩了无数次才AC）

Segments

Submit: 386   Accepted:62

Time Limit: 1000MS  Memory Limit: 65536K

Description
There are some segments on the plane, we assume there 4 possible position concepts for every 2 segments: no common point, countless common points, one common point and the point is also the endpoint of either segment, one common point but the point is not the endpoint of either segment.
译文：平面上有若干线段。线段两两间存在4种可能的位置关系：无公共点，无数公共点，一个公共点且为其中一条的端点，一个公共点但不是端点。

 

 

Input
The first line of the input is an integer T, which represents that there are T test cases.
An integer n indicating the number of segments
n line follows, 4 integers per line, x1, y1, x2, y2, representing the coordinates of the endpoints of both segments .(1<=n<=100)
译文：第一行测试数据组数T；第二行每组中线段数n；接下来n行线段两端点的二维坐标。（n不超过100）

Output
For each test case, please output 4 line, 1 integer per line
For the first line, please output the number of pairs of segments that have no common point
For the second line, please output the number of pairs of segments that have countless common point
For the third line, please output the number of pairs of segments that have 1 common point and the point is also the endpoint of either segment
For the fourth line, please output the number of pairs of segments that have 1 common point and the point is not the endpoint.
After each case output a blank line include the last case。
译文：输出四行，分别是对应四种情况的线段对的数目，每组测试数据后输出一个空行。

Sample Input

2
3
-1 0 1 0
0 -1 0 1
-1 2 1 2
3
1 0 3 0
2 0 4 0
4 0 5 0


Sample Output

2
0
0
1

1
1
1
0


Source
第四届北京邮电大学程序设计竞赛决赛

 

 Code

 /*BOJ 1805 Segmentscoded by [BUPT]-aswmtjdsjtime:2010-07-28@915*//*we get many segmentsthere are four kinds of relationship between these segmentsfirst: no common point (1.parallel 2.not parallel but no intersection)second: have countless point(they get the same part)third: 1 common point and the point is the end point (1.they are connected 2.one's endpoint on the other's middle) (1.1.parallel and connected 1.2.just commonly connected)fourth: 1 common point but not endpoint (intersection and crosspoint isn't endpoint)*//*本题难点在于四种情况有众多子情况，分类不易划分，容易重叠。以及垂直的线段因为斜率的问题又与一般线段不同，需要额外的处理方式。*//*分类：第一种：无公共点1． 平行（即两线段斜率相同或同为正无穷，且线段间距离大于eps）（因为double的精度问题，不能大于0）2． 相离（两条线段不相交，但是所在直线相交——判断线段交点是否在两线段上即可） 第二种：无数公共点 只有重叠这一种情况（斜率相同或同为正无穷的两条线段，距离小于eps，且一条线段的一个端点在另一条线段上（非端点处）） 第三种：（有且仅有）一个公共点且为其中一条的端点1． 两条斜率相同或同为正无穷的线段相连接2． 两条斜率不同的线段相连接3． 一条线段的端点在另一条线段上（非端点位置） 第四种：（有且仅有）一个公共点且非端点（纯朴朴素的相交，交点非各自的端点即可）*//* some knowledge: k * x - y + (y0 - k * x0) = 0 distance between two lines formula : |C1 - C2| / sqrt ( A^2 + B^2)*/#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <cctype>#include <cstdlib>#include <algorithm>#include <vector>#include <ctime>#define no 0#define many 1#define end 2#define cross 3/** time(0)'s return value is* a number approximately near to 1e9*/using namespace std;typedef long long ll;const double eps=1e-7;const ll MAX=1<<30;const double maxk = 99999999999.00;//when segment a is perpendicular to X-axis,its tangent rate become unlimited..we record it as this//冒充斜率正无穷时的斜率typedef struct{ double x1,x2,y1,y2;//maybe it's a little easier by using real numbers to record these points' coordinates //to calculate the tangent rate of the segments double k,l;} segment;//线段结构体，含两端点xy坐标以及斜率和长度（端点可以再用point模板类或者结构体，长度最后没用到）segment seg[105];//segments' initializationint rela[5];//record the number of relation no,many,end,cross and in case of overstack//用于记录线段间关系，正如一开始宏定义，有四种，no,many,end,cross，你懂的typedef struct{ segment a,b; bool paran; double x0,y0; } crosspoint;//crosspoint struct //need initialize all para are true //in order to minimize the calculation quantity of finding crosspoint //first input two segments,by judging parallel relation or not,to calculate the coordinate of crosspoint//交点结构体，含两线段，平行判定（默认平行，主函数中初始化），交点坐标crosspoint cp[105][105];//segment i with j,has crosspoint or not,if does,the coordinate is whatdouble findl(segment t){ return sqrt((t.x1-t.x2)*(t.x1-t.x2)+(t.y1-t.y2)*(t.y1-t.y2)); }//求取线段长度。。没用上double tanr(double x1,double y1,double x2,double y2)//calculate the tangent rate of a segment{ if(fabs(x1-x2) < eps) return maxk; else return (y2-y1)/(x2-x1); }//求取斜率，含正无穷特判bool onseg(double x0,double y0,segment segt)//judge the cross point is on the segment (but not endpoint) or not{ if(fabs(segt.x1-segt.x2) < eps)//parallel to Y-axis { if((y0-segt.y1)*(y0-segt.y2) < eps) return true; } else if(fabs(segt.y1-segt.y2) < eps)//parallel to X-axis { if((x0-segt.x1)*(x0-segt.x2) < eps) return true; } else//general.... { if((x0-segt.x1)*(x0-segt.x2) < eps && (y0-segt.y1)*(y0-segt.y2) < eps) return true; } return false;//else }//判断点在线段上（因为精度的问题，不小心没有排除点在线段端点的情况，所以在主函数中使用了二重判别），含平行于xy轴的特判（因为判别方法的问题。。定比分点坐标公式在垂直或水平的情况下需要特判）bool onpoint(double x0,double y0,segment segt)//judge the crosspoint is on one of the two endpoints of a segment or not{ if(fabs(x0 - segt.x1) < eps && fabs(y0 - segt.y1) < eps|| fabs(x0 - segt.x2) < eps && fabs(y0 - segt.y2) < eps) return true; else return false; }//判断点在线段端点（任一）bool para(segment t1,segment t2)//judge two segments are parallel or not{ if(fabs(t1.k-t2.k) < eps) return true; else return false; }//判断平行crosspoint findcross(segment a,segment b){ crosspoint temp; if(para(a,b)) temp.paran = true; else { temp.paran = false; temp.x0 = (b.y1 - a.y1 + a.k * a.x1 - b.k * b.x1) / ( a.k - b.k); temp.y0 = (temp.x0 - a.x1) * a.k + a.y1; } return temp; }//寻找交点double dis(segment a,segment b){ double A = a.k,B = 1.0,C1 = a.y1 - a.k * a.x1,C2 = b.y1 - a.k * b.x1; return fabs(C1-C2) / sqrt( B + A * A); } //求取线段间距离（主函数中给出前提：平行） int main(){ int t,n,i,j; scanf("%d",&t); while(t--) { for(i=0;i<5;i++)//initialization——初始化 rela[i] = 0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%lf%lf%lf%lf",&seg[i].x1,&seg[i].y1,&seg[i].x2,&seg[i].y2); seg[i].k = tanr(seg[i].x1,seg[i].y1,seg[i].x2,seg[i].y2); seg[i].l = findl(seg[i]); for(j=0;j<n;j++) cp[i][j].paran = true;//if not judged yet,set it as parallel.... }//kinds of initializations//读入及初始化 for(i=0;i<n;i++) { for(j=i+1;j<n;j++) { if(para(seg[i],seg[j]))//平行的情况下（或者说斜率相同的情况下） { if(dis(seg[i],seg[j]) > eps) //相离（真正平行）status:no rela[no]++; else { if(onseg(seg[i].x1,seg[i].y1,seg[j])||onseg(seg[j].x1,seg[j].y1,seg[i])||onseg(seg[j].x2,seg[j].y2,seg[i])||onseg(seg[i].x2,seg[i].y2,seg[j]))//一条线段的端点在另一条线段上 { if(!(onpoint(seg[i].x1,seg[i].y1,seg[j])||onpoint(seg[j].x1,seg[j].y1,seg[i])||onpoint(seg[j].x2,seg[j].y2,seg[i])||onpoint(seg[i].x2,seg[i].y2,seg[j])))//且不在端点上（说明至少部分重合）status:many rela[many]++;// when parallel ,one point on the other(not end),then many points on the other... else//(onpoint(seg[i].x1,seg[i].y1,seg[j])||onpoint(seg[j].x1,seg[j].y1,seg[i])||onpoint(seg[j].x2,seg[j].y2,seg[i])||onpoint(seg[i].x2,seg[i].y2,seg[j]))//在端点上status:end rela[end]++; } else//平行且距离为0但互不相交status:no rela[no]++;//-- ---- } } else//不平行（斜率不同） { cp[i][j] = findcross(seg[i],seg[j]);//寻找交点 if(onpoint(cp[i][j].x0,cp[i][j].y0,seg[i])&&onpoint(cp[i][j].x0,cp[i][j].y0,seg[j]))//remember to modify code after copy&paste//复制粘贴后记得修改应改变的变量名、迭代器名//如果交点为某条线段的端点status：end rela[end]++; else if((onseg(cp[i][j].x0,cp[i][j].y0,seg[i])&&onpoint(cp[i][j].x0,cp[i][j].y0,seg[j]))||(onseg(cp[i][j].x0,cp[i][j].y0,seg[j])&&onpoint(cp[i][j].x0,cp[i][j].y0,seg[i])))//其实重叠了。这个与上一情况重叠。。算我写挫了吧 rela[end]++; else if(onseg(cp[i][j].x0,cp[i][j].y0,seg[i])&&onseg(cp[i][j].x0,cp[i][j].y0,seg[j]))//交点分别在两条线段的非端点位置上status:cross rela[cross]++;//maybe judge wrong because of accuracy lost else//交点不在线段上status:no rela[no]++; } } } printf("%d/n%d/n%d/n%d/n/n",rela[no],rela[many],rela[end],rela[cross]); } //system("pause"); return 0;}