HDU 1086 You can Solve a Geometry Problem too(判断线段相交)

题目地址:HDU 1086

就这么一道仅仅判断线段相交的题目写了2k多B的代码。。是不是有点浪费。。。但是我觉得似乎哪里也优化不了了。。。。

判断线段相交就是利用的叉积。假如现在两条线段分别是L1和L2,先求L1和L2两个端点与L1的某个端点的向量的叉积,如果这两个的叉积的乘积小于0的话,说明L1在是在L2两个端点之间的,但此时并不保证一定相交。此时需要用同样的方法去判断L2是否在L1的两个端点之间,如果L2也在L1的两个端点之间的话,那就足以说明L1与L2相交。但是这题还需要判断是否端点也相交,当时没想到这点,导致白白调了一段时间。。至于端点的判断,我也没想到什么好的方法。。就直接暴力判断4个端点是否是同一点的情况。。

搓代码如下:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
#define eqs 1e-10
struct node
{
    double x, y;
} point[1000];
node xiang(node a, node b)
{
    node f1;
    f1.x=a.x-b.x;
    f1.y=a.y-b.y;
    return f1;
}
int cross(node a, node b)
{
    double c;
    c= a.x*b.y-a.y*b.x;
    if(c>0)
        return 1;
    else if(c==0)
        return 0;
    else
        return -1;
}
int dcmp(double x, double y)
{
    if(fabs(x-y)<=eqs)
        return 1;
    return 0;
}
int main()
{
    int n, i, j;
    int c1, c2, c3, c4, ans;
    while(scanf("%d",&n)!=EOF&&n)
    {
        ans=0;
        for(i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&point[2*i].x,&point[2*i].y,&point[2*i+1].x,&point[2*i+1].y);
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<i; j++)
            {
                c1=cross(xiang(point[2*i],point[2*i+1]),xiang(point[2*j],point[2*i+1]));
                c2=cross(xiang(point[2*i],point[2*i+1]),xiang(point[2*j+1],point[2*i+1]));
                c3=cross(xiang(point[2*j],point[2*j+1]),xiang(point[2*i],point[2*j+1]));
                c4=cross(xiang(point[2*j],point[2*j+1]),xiang(point[2*i+1],point[2*j+1]));
                if(c1*c2<0&&c3*c4<0)
                    ans++;
                else if(dcmp(point[2*i].x,point[2*j].x)&&dcmp(point[2*i].y,point[2*j].y))
                {
                    ans++;
                }
                else if(dcmp(point[2*i].x,point[2*j+1].x)&&dcmp(point[2*i].y,point[2*j+1].y))
                {
                    ans++;
                }
                else if(dcmp(point[2*i+1].x,point[2*j+1].x)&&dcmp(point[2*i+1].y,point[2*j+1].y))
                {
                    ans++;
                }
                else if(dcmp(point[2*i+1].x,point[2*j].x)&&dcmp(point[2*i+1].y,point[2*j].y))
                {
                    ans++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU 1086 You can Solve a Geometry Problem too(判断线段相交),布布扣,bubuko.com

时间: 2024-10-01 13:00:14

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