HDU-1086 You can Solve a Geometry Problem too
2015-05-04 19:28
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#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <algorithm> using namespace std; struct jd { double x1,y1,x2,y2; }; int main() { int n,i,cnt,k,q; double a,b,mi1,ma1,mi2,ma2; struct jd p[105]; while (scanf("%d",&n),n!=0) { k=0;cnt=0; scanf("%lf%lf%lf%lf",&p[k].x1,&p[k].y1,&p[k].x2,&p[k].y2); k++; while (k<n) { q=0; scanf("%lf%lf%lf%lf",&p[k].x1,&p[k].y1,&p[k].x2,&p[k].y2); for (i=0;i<k;i++) { a=((p[k].x1-p[i].x1)*(p[i].y2-p[i].y1)-(p[k].y1-p[i].y1)*(p[i].x2-p[i].x1))* ((p[i].x2-p[i].x1)*(p[k].y2-p[i].y1)-(p[i].y2-p[i].y1)*(p[k].x2-p[i].x1)); b=((p[i].x1-p[k].x1)*(p[k].y2-p[k].y1)-(p[i].y1-p[k].y1)*(p[k].x2-p[k].x1))* ((p[k].x2-p[k].x1)*(p[i].y2-p[k].y1)-(p[k].y2-p[k].y1)*(p[i].x2-p[k].x1)); if (a>=0&&b>=0) cnt++; } k++; } printf("%d\n",cnt); } return 0; }
判断两线段是否相交,可以把两条线段换成向量,如果线段P1P2和直线Q1Q2相交,则P1P2跨立Q1Q2,即:( P1 - Q1 ) × ( Q2 - Q1 ) * ( Q2 - Q1 ) × ( P2 - Q1 ) >= 0,
如果线段Q1Q2和直线P1P2相交,则Q1Q2跨立P1P2,即:( Q1 - P1 ) × ( P2 - P1 ) * ( P2 - P1 ) × ( Q2 - P1 ) >= 0。
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