poj 3449 Geometric Shapes（计算几何）

Geometric Shapes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 1343 Accepted: 565

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 intersectionsB intersects with S, W, and XF intersects with WS intersects with BW intersects with B and FX intersects with BA has no intersectionsB has no intersections

Source

CTU Open 2007

题意：给出一些线段或多边形，问它们是否相交（碰到也不行）

题解：只需把所有输入都化成n边型（线段也是），然后由一个多边形枚举其他多边形，然后再所有边相枚举，求是否相交，而明显线段相交的话2个图形就相交了。

           不过这题的输入输出有点麻烦，有点算是模拟的计算几何了

#include<stdio.h>#include<stdlib.h>#include<string.h>struct point{    double x,y;}mid,my[5];struct polygon{    int n;    char name[18];    struct point a[28];}p[28];char s[10008],ch;int all;double MAX(double x,double y){ return x>y?x:y; }double MIN(double x,double y){ return x<y?x:y; }double cross(struct point p1,struct point p2,struct point p3){    return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);}int cmp(const void *a,const void *b){    struct polygon c=*(struct polygon *)a;    struct polygon d=*(struct polygon *)b;    return strcmp(c.name,d.name);}void add_polygon(int n){    int i;    p[all].n=n;    for(i=0;i<n;i++)    {        scanf("%s",s);        sscanf(s+1,"%lf%c%lf",&p[all].a[i].x,&ch,&p[all].a[i].y);    }}void add_rectangle(){    int i;    p[all].n=4;    for(i=0;i<3;i++)    {        scanf("%s",s);        sscanf(s+1,"%lf%c%lf",&p[all].a[i].x,&ch,&p[all].a[i].y);    }    p[all].a[3].x=p[all].a[0].x+p[all].a[2].x-p[all].a[1].x;    p[all].a[3].y=p[all].a[0].y+p[all].a[2].y-p[all].a[1].y;}void add_square(){    p[all].n=4;    scanf("%s",s);    sscanf(s+1,"%lf%c%lf",&p[all].a[0].x,&ch,&p[all].a[0].y);    scanf("%s",s);    sscanf(s+1,"%lf%c%lf",&p[all].a[2].x,&ch,&p[all].a[2].y);    mid.x=(p[all].a[0].x+p[all].a[2].x)/2;    mid.y=(p[all].a[0].y+p[all].a[2].y)/2;    p[all].a[1].x=mid.x+(p[all].a[0].y-mid.y);    p[all].a[1].y=mid.y-(p[all].a[0].x-mid.x);    p[all].a[3].x=mid.x-(p[all].a[0].y-mid.y);    p[all].a[3].y=mid.y+(p[all].a[0].x-mid.x);}int intersect(struct point p1,struct point p2,struct point p3,struct point p4){    if(MAX(p1.x,p2.x)<MIN(p3.x,p4.x)) return 0;    if(MIN(p1.x,p2.x)>MAX(p3.x,p4.x)) return 0;    if(MAX(p3.y,p4.y)<MIN(p1.y,p2.y)) return 0;    if(MIN(p3.y,p4.y)>MAX(p1.y,p2.y)) return 0;    if(cross(p1,p2,p3)*cross(p1,p2,p4)>1e-8) return 0;    if(cross(p3,p4,p1)*cross(p3,p4,p2)>1e-8) return 0;    return 1;}int check(struct polygon p1,struct polygon p2){    int i,j;    for(i=0;i<p1.n;i++)    {        for(j=0;j<p2.n;j++)            if(intersect(p1.a[i],p1.a[i+1],p2.a[j],p2.a[j+1])) return 1;    }    return 0;}int main(){    int i,j,ans,cou,res[28];    while(scanf("%s",s),s[0]!='.')    {        all=0;        do{            strcpy(p[all].name,s);            scanf("%s",s);            if(s[0]=='l') add_polygon(2);            else if(s[0]=='t') add_polygon(3);            else if(s[0]=='r') add_rectangle();            else if(s[0]=='s') add_square();            else            {                scanf("%d",&i);                add_polygon(i);            }            p[all].a[p[all].n]=p[all].a[0];            all++;        }while(scanf("%s",s),s[0]!='-');        qsort(p,all,sizeof(p[0]),cmp);        for(i=0;i<all;i++)        {            for(ans=j=res[i]=0;j<all;j++)            {                if(i==j) continue;                if(check(p[i],p[j])){ res[j]=1; ans++;}                else res[j]=0;            }            printf("%s ",p[i].name);            if(!ans) printf("has no intersections\n");            else if(ans==1)            {                printf("intersects with ");                for(j=0;j<all;j++) if(res[j]) printf("%s\n",p[j].name);            }            else if(ans==2)            {                printf("intersects with ");                for(j=0;j<all;j++)                {                    if(!res[j]) continue;                    printf("%s and ",p[j].name);                    break;                }                for(j++;j<all;j++) if(res[j]) printf("%s\n",p[j].name);            }            else            {                printf("intersects with");                for(cou=j=0;j<all;j++)                {                    if(!res[j]) continue;                    printf(" %s",p[j].name);                    if(++cou<ans-1) printf(",");                    else if(cou==ans-1) printf(", and");                    else printf("\n");                }            }        }        printf("\n");    }    return 0;}