poj 2826 An Easy Problem?!(线段交，细节题)

【题目大意】：给出两条线段，问这两条线段能够接住多少水。

【解题思路】：先大概估算了一下，这道题一共wa了不下20次。题目的大意不难，难在下面几个注意点上：

1）线段不能平行或者重合

                        2）不能够出现斜率为0的线段（10+wa以上才发现）

                        3）线段相交后开口应该向上

                        4）处于高位的线段的在x轴上的投影不能遮住地位线段的投影（两个方向）

5）求面积的时候，利用叉积/2来求，注意要以地位点的纵坐标为主

【代码】：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

#define eps 1e-8

double p,q;
int cnt;

struct Point
{
double x, y;
Point() {}
Point(double a, double b)
{
x = a, y = b;
}
}point[4];

struct Line
{
Point a, b;
Line() {}
Line(Point x, Point y)
{
a = x, b = y;
}
}line[2];

inline int sig(double k)
{
return k < -eps ? -1 : k > eps;
}

inline double det(double x1, double y1, double x2, double y2)
{
return x1 * y2 - x2 * y1;
}

inline double dotdet(double x1, double y1, double x2, double y2)
{
return x1 * x2 + y1 * y2;
}

inline double dot(Point o, Point a, Point b)
{
return dotdet(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);
}

inline double xmult(Point o, Point a, Point b)
{
return det(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);
}

inline int between(Point o, Point a, Point b)
{
return sig(dot(o, a, b));
}

inline bool sameline(Line u, Line v)
{
if (sig(xmult(u.a, v.a, v.b)) == 0 && sig(xmult(u.b, v.a, v.b)) == 0) return true;
return false;
}

inline bool parallel(Line u, Line v)
{//利用斜率判平行,返回true表示平行
return sig((u.a.x - u.b.x) * (v.a.y - v.b.y) - (v.a.x - v.b.x) * (u.a.y - u.b.y)) == 0;
}

inline int intersectt(Point a, Point b, Point c, Point d, Point &p)
{
double s1, s2, s3, s4;
int d1 = sig(s1 = xmult(a, b, c));
int d2 = sig(s2 = xmult(a, b, d));
int d3 = sig(s3 = xmult(c, d, a));
int d4 = sig(s4 = xmult(c, d, b));
if ((d1^d2) == -2 && (d3^d4) == -2)
{
p.x = (c.x * s2 - d.x * s1) / (s2 - s1);
p.y = (c.y * s2 - d.y * s1) / (s2 - s1);
return 1;
}
if((d1==0&&d2==0)||(d3==0&&d4==0)) return 0;
if (d1 == 0 && between(c, a, b) <= 0) {p=c; return 2;}
if (d2 == 0 && between(d, a, b) <= 0) {p=d; return 2;}
if (d3 == 0 && between(a, c, d) <= 0) {p=a; return 2;}
if (d4 == 0 && between(b, c, d) <= 0) {p=b; return 2;}
return 0;
}

inline int intersect(Line u, Line v, Point &p)
{
return intersectt(u.a, u.b, v.a, v.b, p);
}

inline double get_area(Point a,Point b,Point c)
{
return fabs((a.x*b.y-a.y*b.x+b.x*c.y-c.x*b.y+c.x*a.y-a.x*c.y)/2);
}

double solve(Line a,Line b)
{
if ((point[0].y==point[1].y) ||(point[2].y==point[3].y)) return 0;
if (sameline(a,b)==true) return 0; //判重合
if (parallel(a,b)==true) return 0; //判平行
Point tmp;
int k;
k=intersect(a,b,tmp); //判相交，顺便求交点
if (k==0) return 0;
else
{
//判断开口向上
cnt=0;
for (int i=0; i<4; i++)
if (point[i].y>tmp.y) cnt++;
if (cnt<=1) return 0;
//判断线段的覆盖
if (point[1].x==point[3].x) return 0;
if ((point[1].x<tmp.x&&point[3].x<point[1].x&&xmult(tmp,point[3],point[1])>0)||
(point[1].x>tmp.x&&point[3].x>point[1].x&&xmult(tmp,point[3],point[1])<0))
return 0;
/*
if (point[1].x<=tmp.x && point[3].x<=tmp.x)
{
if (point[1].x>=point[3].x) return 0;
}
//右侧线段的覆盖
if (point[1].x>=tmp.x && point[3].x>=tmp.x)
{
if (point[3].x>=point[1].x) return 0;
}*/
//求面积，以低的木板作为标准
double anss;
Point tmp1;
tmp1=Point((point[1].y-tmp.y)*(point[3].x-tmp.x)/(point[3].y-tmp.y)+tmp.x,point[1].y);
anss=get_area(tmp,point[1],tmp1);
return anss;
}
}

int main()
{
int T;
scanf("%d",&T);
while (T--)
{
scanf("%lf%lf",&p,&q);
point[0]=Point(p,q);
scanf("%lf%lf",&p,&q);
point[1]=Point(p,q);
scanf("%lf%lf",&p,&q);
point[2]=Point(p,q);
scanf("%lf%lf",&p,&q);
point[3]=Point(p,q);
if (point[0].y>point[1].y) swap(point[0],point[1]);
if (point[2].y>point[3].y) swap(point[2],point[3]);
if (point[1].y>point[3].y) {swap(point[0],point[2]); swap(point[1],point[3]); }
line[0]=Line(point[0],point[1]);
line[1]=Line(point[2],point[3]);
printf("%.2lf\n",solve(line[0],line[1]));
}
return 0;
}

【运行结果】wa的太多一个版都放不下。。。。