POJ 3449 Geometric Shapes

Description

While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be cut into special fluorescent materials. To ensure proper processing, the shapes in the picture cannot intersect. However, some logos contain such intersecting shapes. It is necessary to detect them and decide how to change the picture.

Given a set of geometric shapes, you are to determine all of their intersections. Only outlines are considered, if a shape is completely inside another one, it is not counted as an intersection.

Input

Input contains several pictures. Each picture describes at most 26 shapes, each specified on a separate line. The line begins with an uppercase letter that uniquely identifies the shape inside the corresponding picture. Then there is a kind of the shape and two or more points, everything separated by at least one space. Possible shape kinds are:

• square: Followed by two distinct points giving the opposite corners of the square.
• rectangle: Three points are given, there will always be a right angle between the lines connecting the first point with the second and the second with the third.
• line: Specifies a line segment, two distinct end points are given.
• triangle: Three points are given, they are guaranteed not to be co-linear.
• polygon: Followed by an integer number N (3 ≤ N ≤ 20) and N points specifying vertices of the polygon in either clockwise or anti-clockwise order. The polygon will never intersect itself and its sides will have non-zero length.

All points are always given as two integer coordinates X and Y separated with a comma and enclosed in parentheses. You may assume that |X|, |Y | ≤ 10000.

The picture description is terminated by a line containing a single dash (“-”). After the last picture, there is a line with one dot (“.”).

Output

For each picture, output one line for each of the shapes, sorted alphabetically by its identifier (X). The line must be one of the following:

• “X has no intersections”, if X does not intersect with any other shapes.
• “X intersects with A”, if X intersects with exactly 1 other shape.
• “X intersects with A and B”, if X intersects with exactly 2 other shapes.
• “X intersects with A, B, . . ., and Z”, if X intersects with more than 2 other shapes.

Please note that there is an additional comma for more than two intersections. A, B, etc. are all intersecting shapes, sorted alphabetically.

Print one empty line after each picture, including the last one.

Sample Input

A square (1,2) (3,2)
F line (1,3) (4,4)
W triangle (3,5) (5,5) (4,3)
X triangle (7,2) (7,4) (5,3)
S polygon 6 (9,3) (10,3) (10,4) (8,4) (8,1) (10,2)
B rectangle (3,3) (7,5) (8,3)
-
B square (1,1) (2,2)
A square (3,3) (4,4)
-
.

Sample Output

A has no intersections
B intersects with S, W, and X
F intersects with W
S intersects with B
W intersects with B and F
X intersects with B

A has no intersections
B has no intersections

题意是给你一些平面图形，问你每一个图形与哪几个图形有交点。

只用把每个图形的边都处理出来，就变成了判断线段与线段有没有交点的问题了。

只需要注意处理正方形的边：

正方形，已知 (x0,y0) 和(x2,y2)  可以根据下列关系求(x1,y1),(x3,y3)
x1+x3 = x0+x2;
x1-x3  =  y2-y0;
y1+y3 =  y0-y2;
y1-y3 =  x0-x2;

其他看代码就行了。

#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;


const double eps = 1e-8;
const double inf = 1e8;
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);
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
};

struct Line
{
    Point p1,p2;
    Line(){}
    Line(Point _p1,Point _p2)
    {
        p1 = _p1;
        p2 = _p2;
    }
};
struct Shape
{
    char ch;
    char name[10];
    int n;
    Point p[30];
};
bool cmp(Shape a,Shape b)
{
    return a.ch < b.ch;
}
double xmult(Point p0,Point p1,Point p2)
{
    return (p1-p0)^(p2-p0);
}


bool inter(Line a,Line b)
{
    return
        max(a.p1.x,a.p2.x) >= min(b.p1.x,b.p2.x)&&
        max(b.p1.x,b.p2.x) >= min(a.p1.x,a.p2.x)&&
        max(a.p1.y,a.p2.y) >= min(b.p1.y,b.p2.y)&&
        max(b.p1.y,b.p2.y) >= min(a.p1.y,a.p2.y)&&
        xmult(a.p1,b.p1,b.p2)*xmult(a.p2,b.p1,b.p2) < eps&& //小于等于0写成小于eps
        xmult(b.p1,a.p1,a.p2)*xmult(b.p2,a.p1,a.p2) < eps;
}


Shape s[30];

void input(char *str,int i)
{
    if(strcmp(str,"square")==0)
    {
        s[i].n = 4;
        scanf(" (%lf,%lf)",&s[i].p[0].x,&s[i].p[0].y);
        scanf(" (%lf,%lf)",&s[i].p[2].x,&s[i].p[2].y);
        s[i].p[1].x = ((s[i].p[0].x+s[i].p[2].x)+(s[i].p[2].y-s[i].p[0].y))/2;
        s[i].p[1].y = ((s[i].p[0].y+s[i].p[2].y)+(s[i].p[0].x-s[i].p[2].x))/2;
        s[i].p[3].x = ((s[i].p[0].x+s[i].p[2].x)-(s[i].p[2].y-s[i].p[0].y))/2;
        s[i].p[3].y = ((s[i].p[0].y+s[i].p[2].y)-(s[i].p[0].x-s[i].p[2].x))/2;
    }
    else if(strcmp(str,"line")==0)
    {
        s[i].n = 2;
        scanf(" (%lf,%lf)",&s[i].p[0].x,&s[i].p[0].y);
        scanf(" (%lf,%lf)",&s[i].p[1].x,&s[i].p[1].y);
    }
    else if(strcmp(str,"triangle")==0)
    {
        s[i].n = 3;
        scanf(" (%lf,%lf)",&s[i].p[0].x,&s[i].p[0].y);
        scanf(" (%lf,%lf)",&s[i].p[1].x,&s[i].p[1].y);
        scanf(" (%lf,%lf)",&s[i].p[2].x,&s[i].p[2].y);
    }
    else if(strcmp(str,"rectangle")==0)
    {
        s[i].n = 4;
        scanf(" (%lf,%lf)",&s[i].p[0].x,&s[i].p[0].y);
        scanf(" (%lf,%lf)",&s[i].p[1].x,&s[i].p[1].y);
        scanf(" (%lf,%lf)",&s[i].p[2].x,&s[i].p[2].y);
        s[i].p[3].x = s[i].p[2].x + (s[i].p[0].x - s[i].p[1].x);
        s[i].p[3].y = s[i].p[2].y + (s[i].p[0].y - s[i].p[1].y);
    }
    else if(strcmp(str,"polygon")==0)
    {
        scanf("%d",&s[i].n);
        for(int j = 0;j < s[i].n;j++)
        {
            scanf(" (%lf,%lf)",&s[i].p[j].x,&s[i].p[j].y);
        }
    }

}

bool judge(Shape a,Shape b)
{
    for(int i=0;i<a.n;i++)
        for(int j=0;j<b.n;j++)
        {
            if(inter(Line(a.p[i],a.p[(i+1)%a.n]),Line(b.p[j],b.p[(j+1)%b.n])))
                return true;
        }
    return false;
}
int main(void)
{
    int i,j;
    int f[30];
    char str[10];
    while(scanf("%s",str)==1)
    {
        if(str[0] == '.')
            break;
        s[0].ch = str[0];
        scanf("%s",str);
        input(str,0);
        int n = 1;
        while(scanf("%s",str)==1)
        {
            if(str[0] == '-')
                break;
            s[n].ch = str[0];
            scanf("%s",str);
            input(str,n);
            n++;
        }
        sort(s,s+n,cmp);
        for(i=0;i<n;i++)
        {
            memset(f,0,sizeof(f));
            int cnt = 0;
            for(j=0;j<n;j++)
            {
                if(i == j)
                    continue;
                if(judge(s[i],s[j]))
                {
                    cnt++;
                    f[j] = 1;
                }
            }
            printf("%c ",s[i].ch);
            if(cnt == 0)
                printf("has no intersections\n");
            else if(cnt == 1)
            {
                printf("intersects with ");
                for(j=0;j<n;j++)
                    if(f[j] == 1)
                    {
                        printf("%c\n",s[j].ch);
                        break;
                    }
            }
            else if(cnt == 2)
            {
                printf("intersects with ");
                for(j=0;j<n;j++)
                {
                    if(f[j] == 1)
                    {
                        if(cnt != 1)
                            printf("%c",s[j].ch);
                        else
                            printf(" and %c\n",s[j].ch);
                        cnt--;
                    }
                }
            }
            else
            {
                printf("intersects with ");
                for(j=0;j<n;j++)
                {
                    if(f[j] == 1)
                    {
                        if(cnt != 1)
                            printf("%c, ",s[j].ch);
                        else
                            printf("and %c\n",s[j].ch);
                        cnt--;
                    }
                }
            }
        }
        printf("\n");
    }
    return 0;
}