POJ 3449 Geometric Shapes 判断多边形相交

题意不难理解，给出多个多边形，输出多边形间的相交情况（嵌套不算相交），思路也很容易想到。枚举每一个图形再枚举每一条边

恶心在输入输出，不过还好有sscanf()，不懂可以查看cplusplus网站

根据正方形对角的两顶点求另外两个顶点公式：

x2 = (x1+x3-y3+y1)/2; y2 = (x3-x1+y1+y3)/2;

x4= (x1+x3+y3-y1)/2; y4 = (-x3+x1+y1+y3)/2;

还有很多细节要处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1:1;
}

struct Point
{
    double x,y;
    Point() {}
    Point(double _x,double _y)
    {
        x = _x,y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    //叉积
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    //点积
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

struct Line
{
    Point p,q;
    Line() {};
    Line(Point _p,Point _q)
    {
        p = _p,q = _q;
    }
};

//*判断线段相交
bool inter(Line l1,Line l2)
{
    return
        max(l1.p.x,l1.q.x) >= min(l2.p.x,l2.q.x) &&
        max(l2.p.x,l2.q.x) >= min(l1.p.x,l1.q.x) &&
        max(l1.p.y,l1.q.y) >= min(l2.p.y,l2.q.y) &&
        max(l2.p.y,l2.q.y) >= min(l1.p.y,l1.q.y) &&
        sgn((l2.p-l1.q)^(l1.p-l1.q))*sgn((l2.q-l1.q)^(l1.p-l1.q)) <= 0 &&
        sgn((l1.p-l2.q)^(l2.p-l2.q))*sgn((l1.q-l2.q)^(l2.p-l2.q)) <= 0;
}

const int maxn=30;

struct Shape
{
    char id;
    int nump;
    Point p[maxn];
} shp[maxn];

bool cmp(Shape a,Shape b)
{
    return a.id<b.id;
}

char str[20];
int num[maxn];
char inters[maxn][maxn];
map<string,int>mp;

int main()
{
//    freopen("in.txt","r",stdin);
    mp["square"]=2;
    mp["line"]=2;
    mp["triangle"]=3;
    mp["rectangle"]=3;
    int cnt=0;
    while(scanf("%s",str))
    {
        if(strcmp(str,".")==0) break;
        if(strcmp(str,"-")!=0)
        {
            shp[cnt].id=str[0];
            char str_sh[20];
            scanf("%s",str_sh);
            int n;
            if(mp.count(str_sh)) n=mp[str_sh];
            else scanf("%d",&n);
            shp[cnt].nump=0;
            for(int i=0; i<n; i++)
            {
                scanf("%s",str);
                int x,y;
                sscanf(str,"%*c%d%*c%d",&x,&y);
                shp[cnt].p[shp[cnt].nump++]=Point(x,y);
            }
            if(strcmp(str_sh,"rectangle")==0)
            {
                Point p1=shp[cnt].p[0];
                Point p2=shp[cnt].p[1];
                Point p3=shp[cnt].p[2];
                shp[cnt].p[shp[cnt].nump++]=Point(p1.x+p3.x-p2.x,p1.y+p3.y-p2.y);
            }
            if(strcmp(str_sh,"square")==0)
            {
                Point p1=shp[cnt].p[0];
                Point p3=shp[cnt].p[1];
                shp[cnt].p[shp[cnt].nump++]=Point((p1.x+p3.x-p3.y+p1.y)/2,(p3.x-p1.x+p1.y+p3.y)/2);
                shp[cnt].p[shp[cnt].nump++]=Point((p1.x+p3.x+p3.y-p1.y)/2,(-p3.x+p1.x+p1.y+p3.y)/2);
                swap(shp[cnt].p[1],shp[cnt].p[2]);
            }
            cnt++;
            continue;
        }
        sort(shp,shp+cnt,cmp);
        memset(num,0,sizeof(num));
        for(int i=0; i<cnt; i++)
        {
            for(int j=i+1; j<cnt; j++)
            {
                bool flag=false;
                int num1=shp[i].nump;
                int num2=shp[j].nump;
                for(int k1=0; k1<num1 && !flag; k1++)
                {
                    for(int k2=0; k2<num2 && !flag; k2++)
                    {
                        if(inter(Line(shp[i].p[k1],shp[i].p[(k1+1)%num1]),Line(shp[j].p[k2],shp[j].p[(k2+1)%num2])))
                        {
                            inters[i][num[i]++]=shp[j].id;
                            inters[j][num[j]++]=shp[i].id;
                            flag=true;
                        }
                    }
                }
            }
        }
        for(int i=0; i<cnt; i++)
        {
            printf("%c ",shp[i].id);
            if(num[i]==0)
                printf("has no intersections\n");
            else if(num[i]==1)
                printf("intersects with %c\n",inters[i][0]);
            else if(num[i]==2)
                printf("intersects with %c and %c\n",inters[i][0],inters[i][1]);
            else
            {
                printf("intersects with ");
                for(int j=0;j<num[i]-1;j++)
                    printf("%c, ",inters[i][j]);
                printf("and %c\n",inters[i][num[i]-1]);
            }
        }
        cnt=0;
        puts("");
    }
    return 0;
}