POJ 3449 /// 判断线段相交

题目大意：

给出多个多边形及其编号

按编号顺序输出每个多边形与其相交的其他多边形编号

注意一个两个多个的不同输出

将每个多边形处理成多条边 然后去判断与其他多边形的边是否相交

计算正方形另外两点的方法https://blog.csdn.net/qq_33328072/article/details/51655746

根据对角线

x0 + x2 =  x1 + x3

y0 + y2 =  y1 + y3

根据全等三角形

x1 - x3 = y2 - y1

y1 - y3 = x2 - x0

得到

x1 = (x0 + x2 + y2 - y0) / 2

x3 = (x0 + x2 + y0 - y2) / 2

y1 = (y0 + y2 + x0 - x2) / 2

y3 = (y0 + y2 - x0 + x2) / 2

而且注意这里算出了x1后 不能利用x1的结果来计算x3的结果

也就是不能 x3 = x0 + x2 - x1

原因（猜测）可能是x1计算时存在细微的误差

那么用有误差的x1再计算x3的话 误差就更大了 y1y3同理

#include <cstdio>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <map>
#include <vector>
#include <string>
#include <iostream>
using namespace std;

const double eps=1e-;
double add(double a,double b) {
    if(abs(a+b)<eps*(abs(a)+abs(b))) return ;
    return a+b;
}
struct P {
    double x, y;
    P(){};
    P(double _x, double _y):x(_x),y(_y){};
    P operator - (P p) {
        return P(add(x,-p.x),add(y,-p.y)); }
    P operator + (P p) {
        return P(add(x,p.x),add(y,p.y)); }
    P operator * (double d) {
        return P(x*d,y*d); }
    double dot (P p) {
        return add(x*p.x,y*p.y); }
    double det (P p) {
        return add(x*p.y,-y*p.x); }
};
struct L {
    P s,e;
    L(){};
    L(P _s,P _e):s(_s),e(_e){};
};

vector <L> vec[];
vector <int> ans[];

bool onSeg(P a,P b,P c) {
    return (a-c).det(b-c)== && (a-c).dot(b-c)<=;
}
P ins(P a,P b,P c,P d) {
    return a+(b-a)*((d-c).det(c-a)/(d-c).det(b-a));
}
bool insSS(int a,int b)
{
    for(int i=;i<vec[a].size();i++) {
        L s1=vec[a][i];
        for(int j=;j<vec[b].size();j++) {
            L s2=vec[b][j];
            bool flag=;
            if((s1.s-s1.e).det(s2.s-s2.e)==) {
                flag= onSeg(s1.s,s1.e,s2.s) || onSeg(s1.s,s1.e,s2.e)
                   || onSeg(s2.s,s2.e,s1.s) || onSeg(s2.s,s2.e,s1.e);
            }
            else {
                P t=ins(s1.s,s1.e,s2.s,s2.e); //printf("%.2f %.2f\n",t.x,t.y);
                flag= onSeg(s1.s,s1.e,t) && onSeg(s2.s,s2.e,t);
            }
            if(flag) return ;
        }
    }
    return ;
}
void solve()
{
    for(int i=;i<;i++) {
        if(vec[i].size()==) continue;
        for(int j=i+;j<;j++) {
            if(vec[j].size()==) continue;
            if(insSS(i,j)) {
                ans[i].push_back(j);
                ans[j].push_back(i);
            }
        }
    }

    for(int i=;i<;i++) {
        if(vec[i].size()==) continue;
        if(ans[i].size()==) {
            printf("%c has no intersections\n",i+'A');
            continue;
        }
        printf("%c intersects with %c",i+'A',ans[i][]+'A');
        int n=ans[i].size();
        if(n==) {
            printf(" and %c",ans[i][]+'A');
        }
        else {
            for(int j=;j<n;j++) {
                if(j+==n) printf(", and %c",ans[i][j]+'A');
                else printf(", %c",ans[i][j]+'A');
            }
        } printf("\n");
    } printf("\n");
}
void init()
{
    for(int i=;i<;i++)
        vec[i].clear(), ans[i].clear();
}

int main()
{
    string id,ty;
    while(cin>>id) {
        if(id[]=='-') {
            solve(); init(); continue;
        }
        if(id[]=='.') break;
        cin>>ty;
        if(ty=="square") {
            P p0,p1,p2,p3;
            scanf(" (%lf,%lf) (%lf,%lf)",&p0.x,&p0.y,&p2.x,&p2.y);
            p3.x=(p0.x+p2.x-p2.y+p0.y)/2.0; p1.x=(p0.x+p2.x+p2.y-p0.y)/2.0;
            p3.y=(p0.y+p2.y-p0.x+p2.x)/2.0; p1.y=(p0.y+p2.y+p0.x-p2.x)/2.0;
            vec[id[]-'A'].push_back(L(p0,p1));
            vec[id[]-'A'].push_back(L(p2,p1));
            vec[id[]-'A'].push_back(L(p2,p3));
            vec[id[]-'A'].push_back(L(p0,p3));
        }
        else if(ty=="rectangle") {
            P p0,p1,p2,p3;
            scanf(" (%lf,%lf) (%lf,%lf) (%lf,%lf)",&p0.x,&p0.y,&p1.x,&p1.y,&p2.x,&p2.y);
            p3.x=p0.x-p1.x+p2.x; p3.y=p0.y+p2.y-p1.y;
            vec[id[]-'A'].push_back(L(p0,p1));
            vec[id[]-'A'].push_back(L(p2,p1));
            vec[id[]-'A'].push_back(L(p2,p3));
            vec[id[]-'A'].push_back(L(p0,p3));
        }
        else if(ty=="line") {
            P s,e; scanf(" (%lf,%lf) (%lf,%lf)",&s.x,&s.y,&e.x,&e.y);
            vec[id[]-'A'].push_back(L(s,e));
        }
        else if(ty=="triangle") {
            P p0,p1,p2;
            scanf(" (%lf,%lf) (%lf,%lf) (%lf,%lf)",&p0.x,&p0.y,&p1.x,&p1.y,&p2.x,&p2.y);
            vec[id[]-'A'].push_back(L(p0,p1));
            vec[id[]-'A'].push_back(L(p2,p1));
            vec[id[]-'A'].push_back(L(p2,p0));
        }
        else if(ty=="polygon") {
            int n; scanf("%d",&n);
            P a,b; scanf(" (%lf,%lf)",&a.x,&a.y);
            P c=a;
            for(int i=;i<n;i++) {
                b=c;
                scanf(" (%lf,%lf)",&c.x,&c.y);
                vec[id[]-'A'].push_back(L(b,c));
            }
            vec[id[]-'A'].push_back(L(c,a));
        }
    }

    return ;
}