Intersecting Lines
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 12863 | Accepted: 5712 |
Description
We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
Input
The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).
Output
There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".
Sample Input
5 0 0 4 4 0 4 4 0 5 0 7 6 1 0 2 3 5 0 7 6 3 -6 4 -3 2 0 2 27 1 5 18 5 0 3 4 0 1 2 2 5
Sample Output
INTERSECTING LINES OUTPUT POINT 2.00 2.00 NONE LINE POINT 2.00 5.00 POINT 1.07 2.20 END OF OUTPUT
给出n对直线。。。问每对直线的位置关系(平行,重合,相交,相交的话要求出交点)
嗯。
1 /************************************************************************* 2 > File Name: a.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年11月06日 星期五 09时48分39秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #include<cctype> 21 #define fst first 22 #define lson l,m,rt<<1 23 #define rson m+1,r,rt<<1|1 24 #define ms(a,x) memset(a,x,sizeof(a)) 25 using namespace std; 26 const int dx4[4]={1,0,0,-1}; 27 const int dy4[4]={0,-1,1,0}; 28 typedef long long LL; 29 #define sec second 30 const int inf = 0x3f3f3f3f; 31 const double eps=1E-8; 32 int n; 33 34 int dblcmp( double d) 35 { 36 return d<-eps?-1:d>eps; 37 } 38 39 struct point 40 { 41 double x,y; 42 point(){} 43 point(double _x,double _y): 44 x(_x),y(_y){}; 45 void input() 46 { 47 scanf("%lf%lf",&x,&y); 48 } 49 void output() 50 { 51 printf("%.2f %.2f ",x,y); 52 } 53 point sub(point p) 54 { 55 return point(x-p.x,y-p.y); 56 } 57 double det(point p) 58 { 59 return x*p.y-y*p.x; 60 } 61 62 }; 63 64 struct line 65 { 66 point a,b; 67 line(){} 68 void input() 69 { 70 a.input(); 71 b.input(); 72 } 73 74 75 76 int relation(point p) 77 { 78 int c = dblcmp(p.sub(a).det(b.sub(a))); 79 if (c<0) return 1; 80 if (c>0) return 2; 81 return 3; 82 } 83 bool parallel(line v) 84 { 85 return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0; 86 } 87 //0平行 88 //1重合 89 //2相交 90 int linecrossline(line v) 91 { 92 if ((*this).parallel(v)) 93 { 94 return v.relation(a)==3; 95 } 96 return 2; 97 } 98 99 point crosspoint(line v) 100 { 101 double a1 = v.b.sub(v.a).det(a.sub(v.a)); 102 double a2 = v.b.sub(v.a).det(b.sub(v.a)); 103 return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1)); 104 } 105 }li[15][3]; 106 int main() 107 { 108 #ifndef ONLINE_JUDGE 109 freopen("in.txt","r",stdin); 110 #endif 111 scanf("%d",&n); 112 for ( int i = 1 ; i <= n ; i++) 113 { 114 li[i][1].input(); 115 li[i][2].input(); 116 } 117 118 puts("INTERSECTING LINES OUTPUT"); 119 for ( int i = 1 ; i <= n ; i++) 120 { 121 int tmp = li[i][1].linecrossline(li[i][2]); 122 // cout<<"tmp:"<<tmp<<endl; 123 if (tmp==0) 124 { 125 puts("NONE"); 126 } 127 if (tmp==1) 128 { 129 puts("LINE"); 130 } 131 if (tmp==2) 132 { 133 printf("POINT "); 134 li[i][1].crosspoint(li[i][2]).output(); 135 } 136 } 137 puts("END OF OUTPUT"); 138 139 140 141 142 143 #ifndef ONLINE_JUDGE 144 fclose(stdin); 145 #endif 146 return 0; 147 }