poj3449Geometric Shapes

链接

繁琐。

处理出来所有的线段，再判断相交。

对于正方形的已知对角顶点求剩余两顶点 （列出4个方程求解）

p[1].x=(p[0].x+p[2].x+p[2].y-p[0].y)/2;
p[1].y=(p[0].y+p[2].y+p[0].x-p[2].x)/2;
p[3].x=(p[0].x+p[2].x-p[2].y+p[0].y)/2;
p[3].y=(p[0].y+p[2].y-p[0].x+p[2].x)/2;

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define N 600
#define LL long long
#define INF 0xfffffff
#define zero(x) (((x)>0?(x):-(x))<eps)
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
map<string,int>f;
vector<int>ed[30];
int g;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y) {}
} p[600];
typedef point pointt;
pointt operator -(point a,point b)
{
    return pointt(a.x-b.x,a.y-b.y);
}
struct line
{
    pointt u,v;
    int flag;
    char c;
} li[N];
vector<line>dd[30];
char s1[10],s2[15],s[30];
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
point rotate(point a,double rad)
{
    return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double dot(point a,point b)
{
    return a.x*b.x+a.y*b.y;
}
double dis(point a)
{
    return sqrt(dot(a,a));
}
double angle(point a,point b)
{
    return acos(dot(a,b)/dis(a)/dis(b));
}
double cross(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}

double xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
//判三点共线
int dots_inline(point p1,point p2,point p3)
{
    return zero(xmult(p1,p2,p3));
}

//判点是否在线段上,包括端点
int dot_online_in(point p,point l1,point l2)
{
    return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
}

//判两点在线段同侧,点在线段上返回0

int same_side(point p1,point p2,point l1,point l2)
{
    return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;
}

//判两线段相交,包括端点和部分重合

int intersect_in(point u1,point u2,point v1,point v2)
{
    if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2))
        return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);
    return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2);
}
void init(int kk,char c)
{
    int i;
    int  k = c-'A';
    if(kk==1)
    {
        for(i = 0; i <= 2 ; i+=2)
        {
            scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
        }
        p[1].x=(p[0].x+p[2].x+p[2].y-p[0].y)/2;
        p[1].y=(p[0].y+p[2].y+p[0].x-p[2].x)/2;
        p[3].x=(p[0].x+p[2].x-p[2].y+p[0].y)/2;
        p[3].y=(p[0].y+p[2].y-p[0].x+p[2].x)/2;
        p[4] = p[0];
        for(i = 0; i < 4 ; i++)
        {
            li[++g].u = p[i];
            li[g].v = p[i+1];
            dd[k].push_back(li[g]);
        }
    }
    else if(kk==2)
    {
        for(i = 1; i <= 3 ; i++)
        {
            scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
        }
        point pp = point((p[1].x+p[3].x),(p[1].y+p[3].y));
        p[4] = point(pp.x-p[2].x,pp.y-p[2].y);
        //printf("%.3f %.3f
",p[4].x,p[4].y);
        p[5] = p[1];
        for(i = 1; i <= 4; i++)
        {
            li[++g].u = p[i];
            li[g].v = p[i+1];
            li[g].c = c;
            dd[k].push_back(li[g]);
        }
    }
    else if(kk==3)
    {
        for(i = 1; i <= 2; i++)
            scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
        li[++g].u = p[1];
        li[g].v = p[2];
        li[g].c = c;
        dd[k].push_back(li[g]);
    }
    else if(kk==4)
    {
        for(i = 1; i <= 3 ; i++)
            scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
        p[4] = p[1];
        for(i = 1; i <= 3 ; i++)
        {
            li[++g].u = p[i];
            li[g].v = p[i+1];
            li[g].c = c;
            dd[k].push_back(li[g]);
        }
    }
    else if(kk==5)
    {
        int n;
        scanf("%d",&n);
        for(i = 1; i <= n ; i++)
            scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
        p[n+1] = p[1];
        for(i = 1; i <= n ; i++)
        {
            li[++g].u= p[i];
            li[g].v = p[i+1];
            li[g].c = c;
            dd[k].push_back(li[g]);
        }
    }
}

int main()
{
    f["square"] = 1;
    f["rectangle"] = 2;
    f["line"] = 3;
    f["triangle"] = 4;
    f["polygon"] = 5;
    int i,j,k;
    while(scanf("%s",s1)!=EOF)
    {
        if(s1[0]=='.') break;
        if(s1[0]=='-') continue;
        for(i =0 ; i < 26 ; i++)
        {
            ed[i].clear();
            dd[i].clear();
        }
        g = 0;
        k=0;
        scanf("%s",s2);
        s[++k] = s1[0];
        init(f[s2],s1[0]);
        while(scanf("%s",s1)!=EOF)
        {
            if(s1[0]=='-') break;
            //cout<<s1<<endl;
            scanf("%s",s2);
            s[++k] = s1[0];
            init(f[s2],s1[0]);
        }
        //cout<<g<<endl;
        sort(s+1,s+k+1);
        for(i = 1 ; i <= k; i++)
        {
            int u,v;
            u = s[i]-'A';
            //cout<<u<<" "<<dd[u].size()<<endl;
            for(j = i+1; j <= k ; j++)
            {
                v = s[j]-'A';
                int flag = 0;
                for(int ii = 0 ; ii < dd[u].size() ; ii++)
                {
                    for(int jj = 0 ; jj < dd[v].size() ; jj++)
                    {
                        if(intersect_in(dd[u][ii].u,dd[u][ii].v,dd[v][jj].u,dd[v][jj].v))
                        {

                            flag = 1;
                            break;
                        }
//                    if(u==5&&v==22)
//                        {
//                            output(dd[u][ii].u);
//                        output(dd[u][ii].v);
//                        output(dd[v][jj].u);
//                        output(dd[v][jj].v);
//                        }
                    }
                    if(flag) break;
                }
                if(flag)
                {
                    ed[u].push_back(v);
                    ed[v].push_back(u);
                }
            }
        }
        for(i = 1 ; i <= k; i++)
        {
            int u = s[i]-'A';
            if(ed[u].size()==0)
                printf("%c has no intersections
",s[i]);
            else
            {

                sort(ed[u].begin(),ed[u].end());
                if(ed[u].size()==1)
                    printf("%c intersects with %c
",s[i],ed[u][0]+'A');
                else if(ed[u].size()==2)
                    printf("%c intersects with %c and %c
",s[i],ed[u][0]+'A',ed[u][1]+'A');
                else
                {
                    printf("%c intersects with ",s[i]);
                    for(j = 0 ; j < ed[u].size()-1 ; j++)
                        printf("%c, ",ed[u][j]+'A');
                    printf("and %c
",ed[u][j]+'A');
                }
            }
        }
        puts("");
    }
    return 0;
}