poj1410 - Intersection

                                
想看更多的解题报告:http://blog.csdn.net/wangjian8006/article/details/7870410

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题目大意：题目意思很简单，就是说有一个矩阵是实心的，给出一条线段，问线段和矩阵是否相交

解题思路：用到了线段与线段是否交叉，然后再判断线段是否在矩阵里面，这里要注意的是，他给出的矩阵的坐标明显不是左上和右下的坐标，需要自己去判断下左上点与右下点的坐标。

 

/*
Memory 188K
Time    0MS
*/
#include <stdio.h>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a<b?b:a)

typedef struct{
double x1,y1,x2,y2;
}Line;

Line line,line2;
double x1,x2,y1,y2;

double direction(double x,double y,double x1,double y1,double x2,double y2){
double a1=x1-x;
double b1=y1-y;
double a2=x2-x;
double b2=y2-y;
return a1*b2-a2*b1;
}

int on_segment(double x1,double y1,double x2,double y2,double x,double y){
if((min(x1,x2)<=x && x<=max(x1,x2)) && (min(y1,y2)<=y && y<=max(y1,y2)))
return 1;
return 0;
}

int cross(Line v,Line t){
double d1,d2,d3,d4;
d1=direction(t.x1,t.y1,t.x2,t.y2,v.x1,v.y1);
d2=direction(t.x1,t.y1,t.x2,t.y2,v.x2,v.y2);
d3=direction(v.x1,v.y1,v.x2,v.y2,t.x1,t.y1);
d4=direction(v.x1,v.y1,v.x2,v.y2,t.x2,t.y2);
if(d1*d2<0 && d3*d4<0) return 1;
if(!d1 && on_segment(t.x1,t.y1,t.x2,t.y2,v.x1,v.y1)) return 1;
if(!d2 && on_segment(t.x1,t.y1,t.x2,t.y2,v.x2,v.y2)) return 1;
if(!d3 && on_segment(v.x1,v.y1,v.x2,v.y2,t.x1,t.y1)) return 1;
if(!d4 && on_segment(v.x1,v.y1,v.x2,v.y2,t.x2,t.y2)) return 1;
return 0;
}
int main(){
int flag,n;
double tmp;
scanf("%d",&n);
while(n--){
scanf("%lf%lf%lf%lf",&line.x1,&line.y1,&line.x2,&line.y2);
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

if(x1>x2){
tmp=x1,x1=x2,x2=tmp;
}
if(y1<y2){
tmp=y1,y1=y2,y2=tmp;
}

flag=0;
line2.x1=x1;
line2.y1=y1;
line2.x2=x2;
line2.y2=y1;
if(cross(line,line2)) flag=1;
else{
line2.x1=x1;
line2.y1=y1;
line2.x2=x1;
line2.y2=y2;
if(cross(line,line2)) flag=1;
else{
line2.x1=x2;
line2.y1=y1;
line2.x2=x2;
line2.y2=y2;
if(cross(line,line2)) flag=1;
else{
line2.x1=x1;
line2.y1=y2;
line2.x2=x2;
line2.y2=y2;
if(cross(line,line2)) flag=1;
}
}
}
if(flag) printf("T\n");
else{
if(x1<min(line.x1,line.x2) && x2>max(line.x1,line.x2) && y1>max(line.y1,line.y2) && y2<min(line.y1,line.y2))
printf("T\n");
else
printf("F\n");
}
}
return 0;
}

====================================================================

 

/*模拟
Memory 252K
Time    0MS
*/
#include<iostream>

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;

int xstar,ystar,xend,yend;
int xleft,ytop,xright,ybottom;

bool spot(){
if((xstar<=xright && xstar>=xleft && ystar<=ytop && ystar>=ybottom) || (xend<=xright && xend>=xleft && yend<=ytop

&& yend>=ybottom))
return true;
return false;
}

bool line(){
double tmp;
double k;
//考虑斜率不存在的情况
if(xstar==xend && xstar<=xright && xstar>=xleft && ytop<=max(ystar,yend) && ytop>=min(ystar,yend)) return true;
if(xstar==xend) return false;

k=1.0*(ystar-yend)/(xstar-xend);

//与矩形上边相交
tmp=1.0*(ytop-ystar)/k+xstar;
if(ytop<=max(ystar,yend) && ytop>=min(ystar,yend) && tmp<=xright && tmp>=xleft) return true;
//与矩形下边相交
tmp=1.0*(ybottom-ystar)/k+xstar;
if(ybottom<=max(ystar,yend) && ybottom>=min(ystar,yend) && tmp<=xright && tmp>=xleft) return true;
//与矩形左边相交
tmp=1.0*k*(xleft-xstar)+ystar;
if (xleft<=max(xstar,xend) && xleft>=min(xstar,xend) && tmp<=ytop && tmp>=ybottom) return true;
//与矩形右边相交
tmp=1.0*k*(xright-xstar)+ystar;
if (xright<=max(xstar,xend) && xright>=min(xstar,xend) && tmp<=ytop && tmp>=ybottom) return true;

return false;
}

int main(){
int t,tmp;
cin>>t;
while(t--){
cin>>xstar>>ystar>>xend>>yend;
cin>>xleft>>ytop>>xright>>ybottom;

if(xleft>xright){
tmp=xleft;
xleft=xright;
xright=tmp;
}
if(ybottom>ytop){
tmp=ybottom;
ybottom=ytop;
ytop=tmp;
}

if(spot())		//如果端点在矩形的内部
cout<<'T'<<endl;
else if(line())		//如果线段和四条边相交
cout<<'T'<<endl;
else cout<<'F'<<endl;
}
return 0;
}