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POJ1269 直线相交
版子 http://blog.csdn.net/acm_zl/article/details/9471451
#include <iostream> #include <cstdio> #include <string> #include <string.h> #include <map> #include <vector> #include <cstdlib> #include <cmath> #include <algorithm> #include <queue> #include <set> #include <stack> using namespace std; int main() { int t; double x1,y1,x2,y2,x3,y3,x4,y4; scanf("%d", &t); printf("INTERSECTING LINES OUTPUT\n"); while(t--) { scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4); if(x1==x2 && x3==x4)//两直线都没有斜率 { if(x3==x1) printf("LINE\n"); else printf("NONE\n"); } else if(x1==x2 && x3!=x4)//有一条直线斜率存在 { double k = (y4-y3)*1.0/(x4-x3); double b = y3-(k*x3); double ansx = x1; double ansy = k*x1+b; printf("POINT %.2lf %.2lf\n", ansx, ansy); } else if(x1!=x2 && x3==x4)//有一条直线斜率存在 { double k = (y2-y1)*1.0/(x2-x1); double b = y2-(k*x2); double ansx = x3; double ansy = k*x3+b; printf("POINT %.2lf %.2lf\n", ansx, ansy); } else //两条直线斜率都存在 { double k1 = (y2-y1)*1.0/(x2-x1); double b1 = y2-(k1*x2); double k2 = (y4-y3)*1.0/(x4-x3); double b2 = y3-(k2*x3); if(k1==k2) { if(b1==b2) printf("LINE\n"); else printf("NONE\n"); } else { double ansx = (b1-b2)*1.0/(k2-k1); double ansy = k1*ansx + b1; printf("POINT %.2lf %.2lf\n", ansx, ansy); } } } printf("END OF OUTPUT\n"); return 0; }
POJ1269 直线相交
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