POJ 2826 An Easy Problem?! 好的标题

受该两块木板以形成槽的效果。Q槽可容纳雨水多,注意雨爆跌，思想是非常easy，分类讨论是有点差。

1.假定两条线段不相交或平行，然后再装0；

2.有一个平行x轴。连衣裙0。

3.若上面覆盖以下的，装0；

4.其他，叉积求面积。

直接上代码：

#include <iostream>
#include <cmath>
#include <stdio.h>

using namespace std;

const double eps=1e-8;

struct point{
double x;
double y;
};

struct line{
point a;
point b;
}l1,l2;

double ans;

//求叉积
double xmult(point p0 ,point p1 ,point p2){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
int dblcmp(double n){
if(fabs(n)<eps) return 0;
return n>0?1:-1;
}
//推断是否相交。怎样相交
int judge(line l1 ,line l2){
double d1=dblcmp(max(l1.a.x,l1.b.x)-min(l2.a.x,l2.b.x));
double d2=dblcmp(max(l2.a.x,l2.b.x)-min(l1.a.x,l1.b.x));
double d3=dblcmp(max(l1.a.y,l1.b.y)-min(l2.a.y,l2.b.y));
double d4=dblcmp(max(l2.a.y,l2.b.y)-min(l1.a.y,l1.b.y));
double d5=dblcmp(xmult(l2.a,l1.a,l1.b));
double d6=dblcmp(xmult(l2.b,l1.a,l1.b));
double d7=dblcmp(xmult(l1.a,l2.a,l2.b));
double d8=dblcmp(xmult(l1.b,l2.a,l2.b));
if(d1>=0&&d2>=0&&d3>=0&&d4>=0){
if(d5*d6>0||d7*d8>0) return 0;//不相交
else if(d5==0&&d6==0) return 1;//共线相交
else if(d5==0||d6==0||d7==0||d8==0) return 2;//端点相交
else return 3;//规范相交
}
return 0;
}

//求斜率
bool getslope(line l ,double &k){
double t=l.a.x-l.b.x;
if(t==0) return false;
k=(l.a.y-l.b.y)/t;
return true;
}

//求线段交点
point getIntersect(line l1, line l2) {
point p;
double A1 = l1.b.y - l1.a.y;
double B1 = l1.a.x - l1.b.x;
double C1 = (l1.b.x - l1.a.x) * l1.a.y - (l1.b.y - l1.a.y) * l1.a.x;
double A2 = l2.b.y - l2.a.y;
double B2 = l2.a.x - l2.b.x;
double C2 = (l2.b.x - l2.a.x) * l2.a.y - (l2.b.y - l2.a.y) * l2.a.x;
p.x = (C2*B1 - C1*B2) / (A1*B2 - A2*B1);
p.y = (C1*A2 - C2*A1) / (A1*B2 - A2*B1);
if(p.x==-0) p.x=0;
return p;
}

//求a,b两点中y坐标更大的点
point getbiggerY(point a ,point b){
point q;
if (dblcmp(a.y-b.y) > 0) {
q.x = a.x;
q.y = a.y;
} else {
q.x = b.x;
q.y = b.y;
}
return q;
}

double getarea(point p ,point p1 ,point p2){
point q;
double a;
if(dblcmp(p1.y-p2.y)>=0) {
q.y = p2.y;
q.x = p.x+(p1.x-p.x)*(p2.y-p.y) / (p1.y-p.y); //求还有一点的坐标
a=fabs(xmult(p, p2, q)) / 2; //叉积求面积
}
else {
q.y = p1.y;
q.x = p.x+(p2.x-p.x)*(p1.y-p.y) / (p2.y-p.y);
a=fabs(xmult(p, p1, q)) / 2;
}
return a;
}

int main()
{
int t;
cin>>t;
while(t--){
point p1 ,p2 ,p;
cin>>l1.a.x>>l1.a.y>>l1.b.x>>l1.b.y;
cin>>l2.a.x>>l2.a.y>>l2.b.x>>l2.b.y;
if(judge(l1,l2)<=1)//是否相交
ans=0；
else{
p1=getbiggerY(l1.a ,l1.b);
p2=getbiggerY(l2.a ,l2.b);
p=getIntersect(l1 ,l2);
if(!dblcmp(p.x-p1.x)&&!dblcmp(p.y-p1.y) || !dblcmp(p.x-p2.x)&&!dblcmp(p.x-p2.y))//开口方向
ans=0;
else{
double k1 ,k2;
bool f1 ,f2;
f1 = getslope(l1, k1);
f2 = getslope(l2, k2);
if (!dblcmp(k1) || !dblcmp(k2)) ans=0;
else if (f1 && f2) {//假设两条线都不与y轴平行
if (dblcmp(k1*k2) > 0) { //当两条线段的斜率符号同样时。
int d1 = dblcmp(k1-k2);
int d2 = dblcmp(k2);
if (d1>0&&d2>0&&dblcmp(p2.x-p1.x)*dblcmp(p2.x-p.x)<=0
|| d1<0&&d2>0&&dblcmp(p1.x-p2.x)*dblcmp(p1.x-p.x)<= 0
|| d1>0&&d2<0&&dblcmp(p1.x-p2.x)*dblcmp(p1.x-p.x)<= 0
|| d1<0&&d2<0&&dblcmp(p2.x-p1.x)*dblcmp(p2.x-p.x)<= 0)//覆盖情况
ans = 0;
else
ans=getarea(p,p1,p2);
}
else
ans=getarea(p,p1,p2);
}
else
ans=getarea(p,p1,p2);
}
}
printf ("%.2lf\n", ans);
}
return 0;
}