POJ 3449 Geometric Shapes（几何相交）

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

大致题意：给你若干个平面几何的信息，问它们之间的相交关系。如果是线段，则告诉你两个端点的位置，如果是三角形，则告诉你三个端点的位置，如果是正方形，则告诉你对角线上的两个端点的位置，如果是矩形，则告诉你三个端点的位置，如果是n个点的多边形，则告诉你n个端点的位置。

思路：判断两个几何是否相交，只要判断它们的所有线段是否相交即可。然后对于正方形我们还需要求先出它的另外两个点的坐标。
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;

即
x1=(x0+x2+y2-y0)/2;
y1=(x0-x2+y0-y2)/2;
x3=(x0+x2+y0-y2)/2;
y3=(x2-x0+y0-y2)/2;

代码如下

#include<iostream>#include<set>#include<vector>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>using namespace std;typedef long long ll;const double eps=1e-8;int dcmp(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);    }    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 a,b;    Line() {}    Line(Point _a,Point _b) {        a=_a;        b=_b;    }};int inter(Line L1,Line L2)//判断两条线段是否相交,若相交返回1否则0 {    return        max(L1.a.x,L1.b.x) >= min(L2.a.x,L2.b.x) &&        max(L2.a.x,L2.b.x) >= min(L1.a.x,L1.b.x) &&        max(L1.a.y,L1.b.y) >= min(L2.a.y,L2.b.y) &&        max(L2.a.y,L2.b.y) >= min(L1.a.y,L1.b.y) &&        dcmp((L2.a-L1.a)^(L1.b-L1.a))*dcmp((L2.b-L1.a)^(L1.b-L1.a)) <= 0 &&        dcmp((L1.a-L2.a)^(L2.b-L2.a))*dcmp((L1.b-L2.a)^(L2.b-L2.a)) <= 0; }struct Node{    char id;    int num;    Point dian[25];}node[30];bool cmp(Node a,Node b){    return a.id<b.id;}bool check(Node a,Node b){    for(int i=0;i<a.num;i++)    for(int j=0;j<b.num;j++)        if(inter(Line(a.dian[i],a.dian[(i+1)%a.num]),Line(b.dian[j],b.dian[(j+1)%b.num])))        return true;    return false;}int main() {    int tol=0;    char str[100];    while(scanf("%s",str)==1)    {        tol=0;        if(str[0]=='.')break;        node[0].id=str[0];        scanf("%s",str);        if(strcmp(str,"square")==0)        {            node[0].num=4;            scanf(" (%lf,%lf)",&node[0].dian[0].x,&node[0].dian[0].x);            scanf(" (%lf,%lf)",&node[0].dian[2].x,&node[0].dian[2].x);            node[0].dian[1].x = ((node[0].dian[0].x+node[0].dian[2].x)+(node[0].dian[2].y-node[0].dian[0].y))/2;            node[0].dian[1].y = ((node[0].dian[0].y+node[0].dian[2].y)+(node[0].dian[0].x-node[0].dian[2].x))/2;            node[0].dian[3].x = ((node[0].dian[0].x+node[0].dian[2].x)-(node[0].dian[2].y-node[0].dian[0].y))/2;            node[0].dian[3].y = ((node[0].dian[0].y+node[0].dian[2].y)-(node[0].dian[0].x-node[0].dian[2].x))/2;        }        else if(strcmp(str,"line")==0)        {            node[0].num=2;            scanf(" (%lf,%lf)",&node[0].dian[0].x,&node[0].dian[0].y);            scanf(" (%lf,%lf)",&node[0].dian[1].x,&node[0].dian[1].y);        }        else if(strcmp(str,"triangle")==0)        {            node[0].num=3;            scanf(" (%lf,%lf)",&node[0].dian[0].x,&node[0].dian[0].y);            scanf(" (%lf,%lf)",&node[0].dian[1].x,&node[0].dian[1].y);            scanf(" (%lf,%lf)",&node[0].dian[2].x,&node[0].dian[2].y);        }        else if(strcmp(str,"polygon")==0)        {            scanf("%d",&node[0].num);            for(int i=0;i<node[0].num;i++)                scanf(" (%lf,%lf)",&node[0].dian[i].x,&node[0].dian[i].y);        }        else if(strcmp(str,"rectangle")==0)        {            node[0].num=4;            scanf(" (%lf,%lf)",&node[0].dian[0].x,&node[0].dian[0].y);            scanf(" (%lf,%lf)",&node[0].dian[1].x,&node[0].dian[1].y);            scanf(" (%lf,%lf)",&node[0].dian[2].x,&node[0].dian[2].y);            node[0].dian[3].x = node[0].dian[2].x + (node[0].dian[0].x - node[0].dian[1].x);            node[0].dian[3].y = node[0].dian[2].y + (node[0].dian[0].y - node[0].dian[1].y);        }        tol=1;        while(scanf("%s",str)==1)        {            if(str[0]=='-') break;            node[tol].id=str[0];            scanf("%s",str);            if(strcmp(str,"square")==0)            {                node[tol].num=4;                scanf(" (%lf,%lf)",&node[tol].dian[0].x,&node[tol].dian[0].x);                scanf(" (%lf,%lf)",&node[tol].dian[2].x,&node[tol].dian[2].x);                node[tol].dian[1].x = ((node[tol].dian[0].x+node[tol].dian[2].x)+(node[tol].dian[2].y-node[tol].dian[0].y))/2;                node[tol].dian[1].y = ((node[tol].dian[0].y+node[tol].dian[2].y)+(node[tol].dian[0].x-node[tol].dian[2].x))/2;                node[tol].dian[3].x = ((node[tol].dian[0].x+node[tol].dian[2].x)-(node[tol].dian[2].y-node[tol].dian[0].y))/2;                node[tol].dian[3].y = ((node[tol].dian[0].y+node[tol].dian[2].y)-(node[tol].dian[0].x-node[tol].dian[2].x))/2;            }            else if(strcmp(str,"line")==0)            {                node[tol].num=2;                scanf(" (%lf,%lf)",&node[tol].dian[0].x,&node[tol].dian[0].y);                scanf(" (%lf,%lf)",&node[tol].dian[1].x,&node[tol].dian[1].y);            }            else if(strcmp(str,"triangle")==0)            {                node[tol].num=3;                scanf(" (%lf,%lf)",&node[tol].dian[0].x,&node[tol].dian[0].y);                scanf(" (%lf,%lf)",&node[tol].dian[1].x,&node[tol].dian[1].y);                scanf(" (%lf,%lf)",&node[tol].dian[2].x,&node[tol].dian[2].y);            }            else if(strcmp(str,"polygon")==0)            {                scanf("%d",&node[tol].num);                for(int i=0;i<node[tol].num;i++)                    scanf(" (%lf,%lf)",&node[tol].dian[i].x,&node[tol].dian[i].y);            }            else if(strcmp(str,"rectangle")==0)            {                node[tol].num=4;                scanf(" (%lf,%lf)",&node[tol].dian[0].x,&node[tol].dian[0].y);                scanf(" (%lf,%lf)",&node[tol].dian[1].x,&node[tol].dian[1].y);                scanf(" (%lf,%lf)",&node[tol].dian[2].x,&node[tol].dian[2].y);                node[tol].dian[3].x = node[tol].dian[2].x + (node[tol].dian[0].x - node[tol].dian[1].x);                node[tol].dian[3].y = node[tol].dian[2].y + (node[tol].dian[0].y - node[tol].dian[1].y);            }            tol++;        }        sort(node,node+tol,cmp);        int f[30];        for(int i=0;i<tol;i++)        {            printf("%c ",node[i].id);            memset(f,0,sizeof(f));            int cnt=0;            for(int j=0;j<tol;j++)                if(i!=j)                if(check(node[i],node[j]))                {                    f[j]=1;                     cnt++;                }            if(cnt==0)  printf("has no intersections\n");            else if(cnt==1)            {                printf("intersects with ");                for(int j=0;j<tol;j++)                if(f[j])                {                    printf("%c\n",node[j].id);                    break;                }            }            else if(cnt==2)            {                printf("intersects with ");                for(int j=0;j<tol;j++)                if(f[j])                {                    if(cnt==2)  printf("%c ",node[j].id);                    if(cnt==1)                      {                        printf("and %c\n",node[j].id);                        break;                    }                    cnt--;                }            }            else             {                printf("intersects with ");                for(int j=0;j<tol;j++)                if(f[j])                {                    if(cnt>1)   printf("%c, ",node[j].id);                    if(cnt==1)  printf("and %c\n",node[j].id);                    cnt--;                }            }        }    }    return 0;}