POJ 3304 Segment

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直线与线段相交，枚举直线的端点用叉积判断是否有交点

#include "cstring"#include "iostream"#include "algorithm"#include "cstdio"#include "queue"#include "set"#include "cmath"using namespace std;typedef long long LL;const int M=510;const int INF = 0x3f3f3f3f;const double PI = acos(-1.00);const double eps = 10e-8;struct point{    double x,y;};int n;struct node{//    struct point//    {//        double x,y;//    };    point s,t;} s[105];int dis(point a,point b){    double d;    d=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));    if(-eps<=d && d<=eps)          return 0;    return 1;}double chaji(point a,point b,point c)  //x1*y2-x2*y1{    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}double judge(point a,point b){    if(!dis(a,b))        return 0;    for(int i=1; i<=n; i++)  //判断是否有交点        if(chaji(a, b, s[i].s)*chaji(a, b, s[i].t) > eps)            return 0;    return 1;}int main(){    int t;    cin>>t;    while(t--)    {        cin>>n;        for(int i=1; i<=n; ++i)        {            cin>>s[i].s.x>>s[i].s.y>>s[i].t.x>>s[i].t.y;        }        int flag=0;        if(n==1)            flag=1;        for(int i=1; i<n; i++) //枚举线段的端点            for(int j=i+1; j<=n; j++)                if(judge(s[i].s, s[j].s)||judge(s[i].s, s[j].t)||judge(s[i].t, s[j].s)||judge(s[i].t, s[j].t))                {                    flag=1;                    break;                }        cout<<(flag?"Yes!":"No!")<<endl;    }    return 0;}

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