Segments_poj3304_计算几何

Description

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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

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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

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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.

Analysis

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问题可以转换成求是否存在一条直线能穿过所有给出的线段
不难想到这条直线在一定范围内平移旋转仍能保持其原性质(穿过所有的线段)，于是可以将直线平移旋转，使其穿过某些线段的端点
∵两点确定一条直线
∴我们可以枚举两个端点作直线，再枚举线段利用叉积判断是否被直线穿过
poj的pascal编译莫名爆了，可怜某媳妇一秒钟，c++大法好

Code

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#include <stdio.h>using namespace std;struct point{    double x,y;}p[201];double cros(point a,point b,point c){    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);}int main(){    int t,n;    scanf("%d",&t);    while (t--)    {        scanf("%d",&n);        bool flag,ans=false;        int cnt=0;        for (int i=1;i<=n;i++)        {            ++cnt;scanf("%lf%lf",&p[cnt].x,&p[cnt].y);            ++cnt;scanf("%lf%lf",&p[cnt].x,&p[cnt].y);        }        for (int i=1;i<=cnt-1&&!ans;i++)            for (int j=i+1;j<=cnt&&!ans;j++)            {                if (p[i].x==p[j].x&&p[i].y==p[j].y)                    continue;                flag=false;                for (int k=1;k<=cnt;k+=2)                    if (cros(p[i],p[j],p[k])*cros(p[i],p[j],p[k+1])>0)                    {                        flag=true;                        break;                    }                if (!flag)                {                    ans=true;                    printf("Yes!\n");                }            }        if (!ans)            printf("No!\n");    }    return 0;}