Segments(poj3304,判断线段与直线相交)
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http://poj.org/problem?id=3304
Segments
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8270 Accepted: 2507
Description
Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.
Input
Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the two endpoints for one of the segments.
Output
For each test case, your program must output "Yes!", if a line with desired property exists and must output "No!" otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.
Sample Input
3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0
Sample Output
Yes!
Yes!
No!
Source
Amirkabir University of Technology Local Contest 2006
解析:
题意:给出n条线段,判断是否存在这样的一条直线,所有线段的投影在其上,且至少有一个交点
思路:判断直线与线段是否相交:
转化成是否有一直线与所有的线段相交
暴力枚举每两条线段端点形成直线,判断该直线是否与所有线段都相交
注意特殊情况的处理
1.n==1,时这样的直线一定可以
2.枚举端点时注意端点不重合,因为两点才能确定一条直线
192K 32MS C++ 2002B
*/
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;const int maxn=100+10;const double eps=1e-9;int n;struct point{double x,y;point(double x=0,double y=0):x(x),y(y){ }}p[maxn][2];point operator+(point A,point B){return point(A.x+B.x,A.y+B.y);}point operator-(point A,point B){return point(A.x-B.x,A.y-B.y);}point operator/(point A,double p){return point(A.x/p,A.y/p);}point operator*(point A,double p){return point(A.x*p,A.y*p);}bool operator<(const point& A,const point& B){return A.x<B.x||(A.x==B.x&&A.y<B.y);}bool operator==(const point& A,const point& B){return A.x==B.x&&A.y==B.y;}int dcmp(double x){if(fabs(x)<eps)return 0;elsereturn x<0? -1:1;}double dis(point A,point B){return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}double Cross(point A,point B){return A.x*B.y-A.y*B.x;}bool isIntersect(point p1,point p2,point Q1,point Q2){//printf("CROSS=%.0lf\n",Cross(p1-Q1,Q2-Q1)*Cross(Q2-Q1,p2-Q1)); if(Cross(p1-Q1,Q2-Q1)*Cross(Q2-Q1,p2-Q1)>=0) return 1; else return 0;}bool check(point Q1,point Q2){if(dcmp(dis(Q1,Q2))==0)return false;//如果只有一个点是不能构成一条直线for(int i=0;i<n;i++){if(isIntersect(p[i][0],p[i][1],Q1,Q2)==0){return false;}}return true;}int main(){ int i,j,T; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=0;i<n;i++) { scanf("%lf%lf%lf%lf",&p[i][0].x,&p[i][0].y,&p[i][1].x,&p[i][1].y); } int ok=0; if(n==1) ok=1; for(i=0;i<n;i++) { if(ok) break; for(j=0;j<n;j++) { if(check(p[i][0],p[j][0])||check(p[i][1],p[j][0])||check(p[i][0],p[j][1])||check(p[i][1],p[j][1])) { ok=1; //printf("ok==%d\n",ok); break; } } } if(ok) printf("Yes!\n"); else printf("No!\n"); } return 0;}
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