枚举矩形的四个点再枚举矩形的两条竖边
此时一个点与一条竖边与x轴形成唯一一条直线
检验此直线是否通过所有矩形
以下是代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#define inf 1e-8
using namespace std;
int n;
struct rectangle//存储矩形
{
    double x,y;
    double m,n,z;
}data[105];
struct node//存储点与答案
{
    double x,y,z;
}data2[420],ans[105];
struct line//存储竖线
{
    double x,y,m,n,z;
}data3[210];
bool findl(node a,line b,node& temp)//寻找直线temp 若不存在返回false
{
    if(a.z==b.z)
    return false;
    temp.y=b.z*a.y/a.z;
    temp.z=b.z;temp.x=b.x;
    if(temp.y-b.y>-inf && temp.y-b.n<inf)return true;
    return false;
}
bool solve(node a,node b,rectangle c,node& temp)//解决直线是否与矩形相交 不相交返回false
{
    if(a.z==c.z)
    {temp=a;return true;}
    if(b.z==c.z)
    {temp=b;return true;}
    temp.x=(c.z-b.z)*(b.x-a.x)/(b.z-a.z)+b.x;
    temp.y=(c.z-b.z)*(b.y-a.y)/(b.z-a.z)+b.y;
    temp.z=c.z;
    if(temp.x-c.x>-inf && c.m-temp.x>-inf && temp.y-c.y>-inf && c.n-temp.y>-inf)return true;
    return false;
}
int main()
{
    int i,j,k,flag;
    node temp,a;
    //freopen("in.txt","r",stdin);
    while((scanf("%d",&n))!=EOF)
    {
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf%lf",&data[i].x,&data[i].y,&data[i].m,&data[i].n,&data[i].z);
            data3[i*2].x=data2[i*4].x=data[i].x;data3[i*2].y=data2[i*4].y=data[i].y;
            data3[i*2].m=data2[i*4+1].x=data[i].x;data3[i*2].n=data2[i*4+1].y=data[i].n;
            data3[i*2+1].x=data2[i*4+2].x=data[i].m;data3[i*2+1].y=data2[i*4+2].y=data[i].y;
            data3[i*2+1].m=data2[i*4+3].x=data[i].m;data3[i*2+1].n=data2[i*4+3].y=data[i].n;
            data3[i*2].z=data3[i*2+1].z=data2[i*4].z=data2[i*4+1].z=data2[i*4+2].z=data2[i*4+3].z=data[i].z;
        }
        flag=0;
        for(i=0;i<n*4;i++)
        {
            for(j=0;j<n*2;j++)
            {
                if(findl(data2[i],data3[j],temp))
                {
                    a=data2[i];
                    for(k=0;k<n;k++)
                    {
                        if(!solve(a,temp,data[k],ans[k]))
                        break;
                        if(k==n-1){flag=1;break;}
                    }
                }
                if(flag==1)break;
            }
            if(flag==1)break;
        }
        if(flag==1)
        {
            printf("SOLUTION\n");
            printf("%.6f\n",temp.x-(temp.z*(a.x-temp.x)/(a.z-temp.z)));
            for(i=0;i<n;i++)
            {
                printf("%.6f %.6f %.6f\n",ans[i].x,ans[i].y,ans[i].z);//G++要用%f 不用%lf
            }
        }
        else printf("UNSOLVABLE\n");
    }
    return 0;
}