poj3449 Geometric Shapes【计算几何】

含【判断线段相交】、【判断两点在线段两侧】、【判断三点共线】、【判断点在线段上】模板

 

Geometric Shapes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:2105		Accepted: 883

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

Source

CTU Open 2007

题意：

给定一些图形及其坐标（可能是正方形、矩形、三角形、多边形、线段），问每一个图形和其他哪些图形相交

思路：

因为数据量很小，暴力判断每一个图形的每一条边和其他每一个图形的每一条边是否相交

输入输出有点恶心 没别的了

//#include <bits/stdc++.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<stdio.h>
#include<cstring>

using namespace std;
typedef long long int LL;

#define zero(x) (((x) > 0? (x) : -(x)) < eps)
//enum shape{"square", "rectangle", "line", "triangle", "polygon"};
const double eps = 1e-8;
const double pi = 3.141592654;
struct point{
    int x,y;
    point()
    {
        x = 0;
        y = 0;
    }
    point(int _x, int _y)
    {
        x = _x;
        y = _y;
    }
};

struct node{
    int sh;
    int nvertices;
    point p[25];
    char name;
    /*bool operator < (const node &b)const
    {
        name < b.name;
    }*/
};
node shapes[30];

bool cmp(node a, node b)
{
    return a.name < b.name;
}

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);
}

//判断三点共线
int dots_inline(point p1, point p2, point p3)
{
    return zero(xmult(p1, p2, p3));
}

//判断点是否在线段上，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;
}

//判断两点在线段同侧，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;
}

//判断两线段相交，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);

}

/*bool inter(point u1, point u2, point v1, point v2)
{
    return
    max(u1.x, u2.x) >= min(v1.x, v2.x) &&
    max(v1.x, v2.x) >= min(u1.x, u2.x) &&
    max(u1.y, u2.y) >= min(v1.y, v2.y) &&
    max(v1.y, v2.y) >= min(u1.y, u2.y) &&
    sgn((v))
}*/

char str[30];
bool inter[30];
int main()
{
    int cnt = 0;
    while(scanf("%s", str) == 1){
        cnt = 0;
        if(str[0] =='.')break;
        shapes[0].name = str[0];
        scanf("%s", str);
        if(strcmp(str, "square") == 0){
            shapes[0].nvertices = 4;
            int x0, x2, y0, y2;
            //cin>>x0>>y0>>x2>>y2;
            scanf(" (%d,%d)", &x0, &y0);
            scanf(" (%d,%d)", &x2, &y2);
            shapes[0].p[0] = point(x0, y0);
            shapes[0].p[2] = point(x2, y2);
            shapes[0].p[1] = point((x0 + x2 + y2 - y0) / 2, (y0 + y2 + x0 - x2) / 2);
            shapes[0].p[3] = point((x0 + x2 - y2 + y0) / 2, (y0 + y2 - x0 + x2) / 2);
        }
        else if(strcmp(str, "line") == 0){
            shapes[0].nvertices = 2;
            int x0, y0, x1, y1;
            scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1);
            shapes[0].p[0] = point(x0, y0);
            shapes[0].p[1] = point(x1, y1);
        }
        else if(strcmp(str, "triangle") == 0){
            shapes[0].nvertices = 3;
            int x0, y0, x1, y1, x2, y2;
            scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2);
            shapes[0].p[0] = point(x0, y0);
            shapes[0].p[1] = point(x1, y1);
            shapes[0].p[2] = point(x2, y2);
        }
        else if(strcmp(str, "rectangle") == 0){
            shapes[0].nvertices = 4;
            int x0, y0, x1, y1, x2, y2;
            scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2);
            shapes[0].p[0] = point(x0, y0);
            shapes[0].p[1] = point(x1, y1);
            shapes[0].p[2] = point(x2, y2);
            shapes[0].p[3] = point(x2 + x0 - x1, y2 + y0 - y1);
        }
        else if(strcmp(str, "polygon") == 0){
            scanf("%d", &shapes[0].nvertices);
            for(int i = 0; i < shapes[0].nvertices; i++){
                scanf(" (%d,%d)", &shapes[0].p[i].x, &shapes[0].p[i].y);
            }
        }
        cnt++;
        while(scanf("%s", str) == 1){
            if(str[0] == '-')break;
            shapes[cnt].name = str[0];
            scanf("%s", str);
            if(strcmp(str, "square") == 0){
                shapes[cnt].nvertices = 4;
                int x0, x2, y0, y2;
                scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x2, &y2);
                shapes[cnt].p[0] = point(x0, y0);
                shapes[cnt].p[2] = point(x2, y2);
                shapes[cnt].p[1] = point((x0 + x2 + y2 - y0) / 2, (y0 + y2 + x0 - x2) / 2);
                shapes[cnt].p[3] = point((x0 + x2 - y2 + y0) / 2, (y0 + y2 - x0 + x2) / 2);
            }
            else if(strcmp(str, "line") == 0){
                shapes[cnt].nvertices = 2;
                int x0, y0, x1, y1;
                scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1);
                shapes[cnt].p[0] = point(x0, y0);
                shapes[cnt].p[1] = point(x1, y1);
            }
            else if(strcmp(str, "triangle") == 0){
                shapes[cnt].nvertices = 3;
                int x0, y0, x1, y1, x2, y2;
                scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2);
                shapes[cnt].p[0] = point(x0, y0);
                shapes[cnt].p[1] = point(x1, y1);
                shapes[cnt].p[2] = point(x2, y2);
            }
            else if(strcmp(str, "rectangle") == 0){
                shapes[cnt].nvertices = 4;
                int x0, y0, x1, y1, x2, y2;
                scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2);
                shapes[cnt].p[0] = point(x0, y0);
                shapes[cnt].p[1] = point(x1, y1);
                shapes[cnt].p[2] = point(x2, y2);
                shapes[cnt].p[3] = point(x2 + x0 - x1, y2 + y0 - y1);
            }
            else if(strcmp(str, "polygon") == 0){
                scanf("%d", &shapes[cnt].nvertices);
                for(int i = 0; i < shapes[cnt].nvertices; i++){
                    scanf(" (%d,%d)", &shapes[cnt].p[i].x, &shapes[cnt].p[i].y);
                }
            }
            cnt++;
        }

        sort(shapes, shapes + cnt, cmp);
        for(int i = 0; i < cnt; i++){
            printf("%c ", shapes[i].name);
            memset(inter, 0, sizeof(inter));
            int ans = 0;
            for(int j = 0; j < cnt; j++){
                if(i == j)continue;
                for(int a = 0; a < shapes[i].nvertices; a++){
                    for(int b = 0; b < shapes[j].nvertices; b++){
                        if(intersect_in(shapes[i].p[a], shapes[i].p[(a + 1) % shapes[i].nvertices], shapes[j].p[b], shapes[j].p[(b + 1) % shapes[j].nvertices])){
                            inter[j] = true;
                        }
                    }
                }
                if(inter[j] == true){
                    ans++;
                }
            }
            if(ans == 0){
                printf("has no intersections
");
            }
            else{
                printf("intersects with");
                if(ans == 1){
                    for(int j = 0; j < cnt; j++){
                        if(inter[j]){
                            printf(" %c
", shapes[j].name);
                        }
                    }
                }
                else if(ans == 2){
                    int tmp = 0;
                    for(int j = 0; j < cnt; j++){
                        if(inter[j]){
                            if(tmp){
                                printf(" and");
                            }
                            printf(" %c", shapes[j].name);
                            tmp++;
                        }
                    }
                    printf("
");
                }
                else{
                    int tmp = 0;
                    for(int j = 0; j < cnt; j++){
                        if(inter[j]){
                            if(tmp == ans - 1){
                                printf(" and %c", shapes[j].name);
                            }
                            else{
                                printf(" %c,", shapes[j].name);
                            }
                            tmp++;
                        }

                    }
                    printf("
");
                }
            }
        }
        printf("
");
    }
    return 0;
}