An Easy Problem?!

题目链接：https://vjudge.net/problem/POJ-2826#author=0

题意：给你两个线段板子，用这个接雨水，问最多能接多少水。

思路：wa了13发终于过了，少考虑了一种情况一直wa。首先如果有一块板子是水平的，那肯定接不了雨水；其次，如果二块板子没有交点，那接的雨水都会漏掉；最后如果两块板子有交点且那个交点不是某块板子的最高点，然后要特判一种特殊情况，它是接不到雨水的，特判图形如下。然后接的水的高度是较低的那块板子。这些找线段的交点和一些判断什么的都可以用差积解决。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
#define INF 0x7ffffff
double chaj(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        double x1,y1,x2,y2;
        double x3,y3,x4,y4;
        cin>>x1>>y1>>x2>>y2;
        cin>>x3>>y3>>x4>>y4;
        if(y1==y2||y3==y4)
        {
            cout<<"0.00"<<endl;
            continue;
        }
        if(chaj(x3-x1,y3-y1,x2-x1,y2-y1)==0&&chaj(x4-x1,y4-y1,x2-x1,y2-y1)==0)
        {
            cout<<"0.00"<<endl;
            continue;
        }
        if(chaj(x3-x1,y3-y1,x2-x1,y2-y1)*chaj(x4-x1,y4-y1,x2-x1,y2-y1)<=0&&chaj(x1-x3,y1-y3,x4-x3,y4-y3)*chaj(x2-x3,y2-y3,x4-x3,y4-y3)<=0)
        {
            double s1=fabs(chaj(x3-x1,y3-y1,x4-x1,y4-y1));
            double s2=fabs(chaj(x3-x2,y3-y2,x4-x2,y4-y2));
            double w=s1/(s1+s2);
            double x=w*(x2-x1)+x1,y=w*(y2-y1)+y1;
            double x5,y5,x6,y6;
            if(y1>y2)
            {
                x5=x1;
                y5=y1;
            }
            else
            {
                x5=x2;
                y5=y2;
            }
            if(y3>y4)
            {
                x6=x3;
                y6=y3;
            }
            else
            {
                x6=x4;
                y6=y4;
            }
            if(y5<=y||y6<=y)
            {
                cout<<"0.00"<<endl;
                continue;
            }
            x5-=x;y5-=y;
            x6-=x;y6-=y;
            if(x5<0&&x6<0)
            {
                if(y5<y6&&x5>=x6&&chaj(x5,y5,x6,y6)<0)
                {
                    cout<<"0.00"<<endl;
                    continue;
                }
                if(y6<y5&&x6>=x5&&chaj(x6,y6,x5,y5)<0)
                {
                    cout<<"0.00"<<endl;
                    continue;
                }
            }
            if(x5>0&&x6>0)
            {
                if(y5<y6&&x5<=x6&&chaj(x5,y5,x6,y6)>0)
                {
                    cout<<"0.00"<<endl;
                    continue;
                }
                if(y5>y6&&x5>=x6&&chaj(x6,y6,x5,y5)>0)
                {
                    cout<<"0.00"<<endl;
                    continue;
                }
            }
            if(y5>y6)
            {
                x5=y6*x5/y5;
                y5=y6;
            }
            else if(y5<y6)
            {
                x6=y5*x6/y6;
                y6=y5;
            }
            double ans=fabs(chaj(x5,y5,x6,y6));
            printf("%.2lf
",ans/2);
        }
        else
            cout<<"0.00"<<endl;
    }
}