poj 3304 Segments(判断直线与线段相交)

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6237		Accepted: 1828

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

题目： http://poj.org/problem?id=3304

题意：给你n条线段，求是否存在这样一条直线，使得所有线段在直线上的投影有公共部分，至少是一个点

分析：根据题目所求可以转换为求是否存在一条直线与所有线段都相交，这样就可以简单的枚举所有线段的端点，构成的直线，判断该直线是否跟所有线段相交即可，主要是直线与线段相交的模板，呵呵

模板写成这样貌似没有以前精简，不过有利于拓展= =

代码：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
const int mm=222;
typedef double mType;
struct Tpoint
{
    mType x,y;
    Tpoint(){}
    Tpoint(mType _x,mType _y):x(_x),y(_y){}
};
struct Tsegment
{
    Tpoint a,b;
    Tsegment(){}
    Tsegment(Tpoint _a,Tpoint _b):a(_a),b(_b){}
    Tsegment(mType sx,mType sy,mType tx,mType ty):a(sx,sy),b(tx,ty){}
};
Tpoint MakeVector(Tpoint P,Tpoint Q)
{
    Tpoint tmp=Tpoint(Q.x-P.x,Q.y-P.y);
    return tmp;
}
mType CrossProduct(Tpoint P,Tpoint Q)
{
    return P.x*Q.y-P.y*Q.x;
}
mType MultiCross(Tpoint P,Tpoint Q,Tpoint R)
{
    return CrossProduct(MakeVector(Q,P),MakeVector(Q,R));
}
bool IsLineIntersect(Tsegment P,Tsegment Q)
{
    return (MultiCross(P.b,P.a,Q.a)*MultiCross(P.b,P.a,Q.b)<=0);
}
int i,n,t;
double sx,sy,tx,ty;
Tsegment g[mm];
int count(Tsegment P)
{
    if(fabs(P.a.x-P.b.x)<1e-8&&fabs(P.a.y-P.b.y)<1e-8)return 0;
    int i,ret=0;
    for(i=0;i<n;++i)
        if(IsLineIntersect(P,g[i]))++ret;
    return ret;
}
bool ok()
{
    int i,j;
    Tsegment tmp;
    for(i=0;i<n;++i)
        for(j=i;j<n;++j)
        {
            tmp=Tsegment(g[i].a,g[j].a);
            if(count(tmp)>=n)return 1;
            tmp=Tsegment(g[i].a,g[j].b);
            if(count(tmp)>=n)return 1;
            tmp=Tsegment(g[i].b,g[j].a);
            if(count(tmp)>=n)return 1;
            tmp=Tsegment(g[i].b,g[j].b);
            if(count(tmp)>=n)return 1;
        }
    return 0;
}
void test()
{
    Tsegment tmp=Tsegment(0,0,1,1);
    printf("%.2lf %.2lf %.2lf %.2lf\n",tmp.a.x,tmp.a.y,tmp.b.x,tmp.b.y);
    if(IsIntersect(Tsegment(0,0,0.5,0.499),Tsegment(1,0,0,1)))puts("Yes");else puts("No");
}
int main()
{
    //test();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;++i)
        {
            scanf("%lf%lf%lf%lf",&sx,&sy,&tx,&ty);
            g[i]=Tsegment(sx,sy,tx,ty);
        }
        printf(ok()?"Yes!\n":"No!\n");
    }
    return 0;
}