poj 1410

给定一个矩形和一条线段，求线段能否交矩形，线段与矩形的边或矩形内部有交点即为交。很简单的一道判断线段相交的题目，不过要正确理解题目意思：相交要包括矩形的内部，即线段在矩形内部也是True。（The rectangle consists of four straight lines and the area in between.）题目说的左上、右下并非座标中的位置，数据给的点也会是左下、右上！！！（The terms top left and bottom right do not imply any ordering of coordinates. ）刚开始写的代码，一直错，后来终于发现，判断线段相交的模板有问题，亏我用这个模板ac了好几道题，通过这道题发现原来我写的模板是有问题的。/*9 1 9 2 4 3 9 69 2 9 1 4 3 9 6F F */ 这组测试样例通不过#include<stdio.h>struct node{double x,y;};struct pp{node a,b;}line[10];double multi(node a,node b,node c){return ((a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x));}double min(double a,double b){return a<b?a:b;}double max(double a,double b){return a>b?a:b;}int main(){double n1,n2,n3,n4,t1,t2,t3,t4,a,b,c,d;int n,i,j;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%lf%lf%lf%lf",&line[0].a.x,&line[0].a.y,&line[0].b.x,&line[0].b.y);scanf("%lf%lf%lf%lf",&a,&b,&c,&d);n1=max(a,c);n2=min(a,c);n3=max(b,d);n4=min(b,d);line[1].a.x=n2;line[1].a.y=n4;line[1].b.x=n2;line[1].b.y=n3;line[2].a.x=n2;line[2].a.y=n3;line[2].b.x=n1;line[2].b.y=n3;line[3].a.x=n1;line[3].a.y=n3;line[3].b.x=n1;line[3].b.y=n4;line[4].a.x=n1;line[4].a.y=n4;line[4].b.x=n2;line[4].b.y=n4;for(j=1;j<=4;j++){t1=multi(line[j].a,line[j].b,line[0].a);t2=multi(line[j].a,line[j].b,line[0].b);t3=multi(line[0].a,line[0].b,line[j].a);t4=multi(line[0].a,line[0].b,line[j].b);if(t1*t2<=0&&t3*t4<=0)break;}if(j<=4)printf("T\n");else{a=line[0].a.x;b=line[0].a.y;c=line[0].b.x;d=line[0].b.y;if(a>n2&&a<n1&&c>n2&&c<n1&&b>n4&&b<n3&&d>n4&&d<n3)printf("T\n");elseprintf("F\n");}}return 0;}正确代码：#include<stdio.h>#include<math.h>struct node{double x,y;};struct pp{node a,b;}line[10];double min(double a,double b){return a<b?a:b;}double max(double a,double b){return a>b?a:b;}double multi(node p0, node p1, node p2){    return ( p1.x - p0.x )*( p2.y - p0.y )-( p2.x - p0.x )*( p1.y - p0.y );}//相交返回true,否则为false, 接口为两线段的端点bool isIntersected(node s1,node e1, node s2,node e2){    return  (max(s1.x,e1.x) >= min(s2.x,e2.x))  &&            (max(s2.x,e2.x) >= min(s1.x,e1.x))  &&            (max(s1.y,e1.y) >= min(s2.y,e2.y))  &&            (max(s2.y,e2.y) >= min(s1.y,e1.y))  &&            (multi(s1,e1,s2)*multi(s1,e1,e2)<=0) &&            (multi(s2,e2,s1)*multi(s2,e2,e1)<=0);}int main(){double n1,n2,n3,n4,a,b,c,d;int n,i,j;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%lf%lf%lf%lf",&line[0].a.x,&line[0].a.y,&line[0].b.x,&line[0].b.y);scanf("%lf%lf%lf%lf",&a,&b,&c,&d);n1=max(a,c);n2=min(a,c);n3=max(b,d);n4=min(b,d);line[1].a.x=n2;line[1].a.y=n4;line[1].b.x=n2;line[1].b.y=n3;line[2].a.x=n2;line[2].a.y=n3;line[2].b.x=n1;line[2].b.y=n3;line[3].a.x=n1;line[3].a.y=n3;line[3].b.x=n1;line[3].b.y=n4;line[4].a.x=n1;line[4].a.y=n4;line[4].b.x=n2;line[4].b.y=n4;for(j=1;j<=4;j++){if(isIntersected(line[0].a,line[0].b,line[j].a,line[j].b))break;}if(j<=4)printf("T\n");else{a=line[0].a.x;b=line[0].a.y;c=line[0].b.x;d=line[0].b.y;if(a>n2&&a<n1&&c>n2&&c<n1&&b>n4&&b<n3&&d>n4&&d<n3)printf("T\n");elseprintf("F\n");}}return 0;}测试数据： 684 9 11 2 1 1 7 511 2 4 9 1 1 7 512 12 24 24 19 5 25 174 6 15 9 1 1 11 1119 5 25 17 12 12 24 240 18 8 12 1 1 11 112 4 4 2 1 1 11 11-4 9 -11 2 -1 1 -7 5-11 2 -4 9 -1 1 -7 5-12 12 -24 24 -19 5 -25 17-4 6 -15 9 -1 1 -11 11-19 5 -25 17 -12 12 -24 240 18 -8 12 -1 1 -11 11-2 4 -4 2 -1 1 -11 114 -9 11 -2 1 -1 7 -511 -2 4 -9 1 -1 7 -512 -12 24 -24 19 -5 25 -174 -6 15 -9 1 -1 11 -1119 -5 25 -17 12 -12 24 -240 -18 8 -12 1 -1 11 -112 -4 4 -2 1 -1 11 -11-4 -9 -11 -2 -1 -1 -7 -5-11 -2 -4 -9 -1 -1 -7 -5-12 -12 -24 -24 -19 -5 -25 -17-4 -6 -15 -9 -1 -1 -11 -11-19 -5 -25 -17 -12 -12 -24 -240 -18 -8 -12 -1 -1 -11 -11-2 -4 -4 -2 -1 -1 -11 -119 1 9 2 4 3 9 69 2 9 1 4 3 9 610 3 13 3 4 3 9 613 3 10 3 4 3 9 610 6 14 6 4 3 9 614 6 10 6 4 3 9 69 7 9 10 4 3 9 69 10 9 7 4 3 9 64 7 4 10 4 3 9 64 10 4 7 4 3 9 60 6 3 6 4 3 9 63 6 0 6 4 3 9 61 3 3 3 4 3 9 63 3 1 3 4 3 9 64 0 4 2 4 3 9 64 2 4 0 4 3 9 65 3 8 5 4 3 9 68 5 5 3 4 3 9 65 3 8 3 4 3 9 68 3 5 3 4 3 9 66 4 6 5 4 3 9 66 5 6 4 4 3 9 64 3 9 6 4 3 9 69 6 4 3 4 3 9 64 3 5 4 4 3 9 65 4 4 3 4 3 9 65 3 8 3 4 3 9 68 3 5 3 4 3 9 65 3 9 3 4 3 9 69 3 5 3 4 3 9 64 4 4 5 4 3 9 64 5 4 4 4 3 9 64 3 4 5 4 3 9 64 5 4 3 4 3 9 64 3 4 6 4 3 9 64 6 4 3 4 3 9 69 2 9 5 4 3 9 69 5 9 2 4 3 9 69 2 9 7 4 3 9 69 7 9 2 4 3 9 6outFFFTTFTFFFTTFTFFFTTFTFFFTTFTFFFFFFFFFFFFFFFFTTTTTTTTTTTTTTTTTTTTTTTT