POJ 3304 直线与线段相交判断

来源：互联网发布：什么是软件集成编辑：程序博客网时间：2024/05/29 17:23
code:
#include <iostream>#include <cstring>#include <cmath>using namespace std;typedef double ld;const ld eps = 1e-8;const int N = 110;struct line{    ld p[2][2];}l[N];struct vec{    ld x, y;    vec(ld _x, ld _y):x(_x), y(_y){};    ld operator * (const vec &rhs) const{        return x * rhs.y - y * rhs.x;    }/**叉积*/};int n;bool f;bool check(ld sx, ld sy, ld ex, ld ey, int ii, int ij){    if(fabs(ex - sx) < eps && fabs(ey - sy) < eps) return false;/**重点*/    for(int i = 1; i <= n; ++i){        if(i == ii || i == ij) continue;        ld la = vec(l[i].p[0][0] - sx, l[i].p[0][1] - sy) * vec(ex - sx, ey - sy);        ld ra = vec(l[i].p[1][0] - sx, l[i].p[1][1] - sy) * vec(ex - sx, ey - sy);        if(la * ra > 0) return false;/**直线和线段相交la * ra <= 0*/    }    return true;}void solve(){    for(int i = 1; i <= n; ++i){        for(int ii = 0; ii < 2; ++ii)/**枚举端点*/            for(int j = 1; j <= n; ++j){                //if(j == i) continue;                for(int ij = 0; ij < 2; ++ij){                    if(check(l[i].p[ii][0], l[i].p[ii][1], l[j].p[ij][0], l[j].p[ij][1], i, j)){                        f = 1;                        break;                    }                }            }    }}int main(){    int t;    cin >> t;    while(t--){        f = 0;        cin >> n;        for(int i = 1; i <= n; ++i) cin >> l[i].p[0][0] >> l[i].p[0][1] >> l[i].p[1][0] >> l[i].p[1][1];        solve();        if(f) cout << "Yes!" << endl;        else cout << "No!" << endl;    }    return 0;}
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