zoj 1648 Circuit Board

计算几何线段相交问题，第一次写，所以没有用模版，可以先参考一下这计算几何算法概览

不建议直接套模版，还是先理解一下
过几天将计算几何专题整理一下，再搞模版

/* ***********************************************
Author        :xryz
Email         :523689985@qq.com
Created Time  :3-31 21:09:06
File Name     :\Users\xryz\Desktop\CircuitBoard.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
int main()
{   int i,j,n;
    double x1[2024],x2[2024],y1[2024],y2[2024],t1,t2,t3,t4;
    bool f;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%lf%lf%lf%lf",&x1[i],&y1[i],&x2[i],&y2[i]);
        f=1;
        for(i=0;i<n-1;i++)//判断线段是否分别跨立
        {
            for(j=i+1;j<n;j++)
            {
                t1=((x1[i]-x1[j])*(y2[j]-y1[j])-(y1[i]-y1[j])*(x2[j]-x1[j]))*
                    ((x2[j]-x1[j])*(y2[i]-y1[j])-(y2[j]-y1[j])*(x2[i]-x1[j]));
                if(t1>0)
                {
                    t2=((x1[j]-x1[i])*(y2[i]-y1[i])-(y1[j]-y1[i])*(x2[i]-x1[i]))*
                        ((x2[i]-x1[i])*(y2[j]-y1[i])-(y2[i]-y1[i])*(x2[j]-x1[i]));
                    if(t2>0) {f=0;break;}
                }

            }
            if(!f) break;
        }
        if(f) printf("ok!\n");
        else printf("burned!\n");
    }
    return 0;
}

#include <stdio.h>
struct Point
{
    double x,y;
};

Point a[2048],b[2048];

Point operator - (Point m,Point n)
{
    Point c;
    c.x=m.x-n.x;
    c.y=m.y-n.y;
    return c;
}

double cross(Point m,Point n)
{
    return m.x*n.y-n.x*m.y;
}

int main()
{
    int i,j,n;
    int flag;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&b[i].x,&b[i].y);
        flag=1;
        for(i=0;i<n-1;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(cross(a[i]-a[j],b[j]-a[j])*cross(b[j]-a[j],b[i]-a[j])>0)
                {
                    if(cross(a[j]-a[i],b[i]-a[i])*cross(b[i]-a[i],b[j]-a[i])>0)
                        flag=0;
                }
                if(flag==0) break;
            }
            if(flag==0) break;
        }

        if(flag) printf("ok!\n");
        else printf("burned!\n");
    }
    return 0;
}

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