POJ 3449 Geometric Shapes

POJ_3449

Because there are very few images, the idea is still very direct. It is enough to enumerate any two images and enumerate the positional relationship between all the line segments of the two images. If there are two line segments intersection, the two figures are intersecting.

The input and output of this question are disgusting. In addition, we can use the vector rotation to obtain the coordinates of the center of the square for a pair of points known to the square, then, we regard the link from the center to the two points as two vectors. By rotating these two vectors 90 degrees each, we can get the vector from the center to the other two points, then we can obtain the coordinates of the other two points. For details about vector rotation, see http://www.matrix67.com/blog/archives/276 /.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 30
#define zero 1e-8
char b[110];
int N, name[MAXD], vn[MAXD], g[MAXD][MAXD], n[MAXD], vis[MAXD];
double x[MAXD][MAXD], y[MAXD][MAXD];
int cmp(const void *_p, const void *_q)
{
int *p = (int *)_p;
int *q = (int *)_q;
return *p - *q;
}
void input()
{
int i, j, k;
    name[N] = b[0];
    scanf("%s", b);
if(strcmp(b, "square") == 0)
    {
        vn[N] = 4;
        scanf("%s", b);
        sscanf(b, "(%lf,%lf)", &x[N][0], &y[N][0]);
        scanf("%s", b);
        sscanf(b, "(%lf,%lf)", &x[N][2], &y[N][2]);
double dx, dy, x0, y0;
        x0 = (x[N][0] + x[N][2]) / 2, y0 = (y[N][0] + y[N][2]) / 2;
        dx = x[N][0] - x0, dy = y[N][0] - y0;
        x[N][1] = x0 - dy, y[N][1] = y0 + dx;
        dx = x[N][2] - x0, dy = y[N][2] - y0;
        x[N][3] = x0 - dy, y[N][3] = y0 + dx;
        x[N][4] = x[N][0], y[N][4] = y[N][0];
    }
else if(strcmp(b, "rectangle") == 0)
    {
        vn[N] = 4;
for(i = 0; i < 3; i ++)
        {
            scanf("%s", b);
            sscanf(b, "(%lf,%lf)", &x[N][i], &y[N][i]);
        }
        x[N][3] = x[N][0] + (x[N][2] - x[N][1]), y[N][3] = y[N][0] + (y[N][2] - y[N][1]);
        x[N][4] = x[N][0], y[N][4] = y[N][0];
    }
else if(strcmp(b, "line") == 0)
    {
        vn[N] = 1;
for(i = 0; i < 2; i ++)
        {
            scanf("%s", b);
            sscanf(b, "(%lf,%lf)", &x[N][i], &y[N][i]);
        }
    }
else if(strcmp(b, "triangle") == 0)
    {
        vn[N] = 3;
for(i = 0; i < 3; i ++)
        {
            scanf("%s", b);
            sscanf(b, "(%lf,%lf)", &x[N][i], &y[N][i]);
        }
        x[N][3] = x[N][0], y[N][3] = y[N][0];
    }
else
    {
        scanf("%d", &vn[N]);
for(i = 0; i < vn[N]; i ++)
        {
            scanf("%s", b);
            sscanf(b, "(%lf,%lf)", &x[N][i], &y[N][i]);
        }
        x[N][vn[N]] = x[N][0], y[N][vn[N]] = y[N][0];
    }
}
int init()
{
int i, j, k;
for(N = 0; ; N ++)
    {
        scanf("%s", b);
if(b[0] == '.')
return 0;
if(b[0] == '-')
break;
        input();
    }
return 1;
}
double fabs(double x)
{
return x < 0 ? -x : x;
}
double dcmp(double x)
{
if(fabs(x) < zero)
return 0;
if(x < 0)
return -1;
return 1;
}
double det(double x1, double y1, double x2, double y2)
{
return x1 * y2 - x2 * y1;
}
int inseg(double x, double y, double x1, double y1, double x2, double y2)
{
if(fabs(x1 - x2) < fabs(y1 - y2))
    {
if(dcmp((y1 - y) * (y2 - y)) <= 0)
return 1;
    }
else
    {
if(dcmp((x1 - x) * (x2 - x)) <= 0)
return 1;
    }
return 0;
}
int check(double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4)
{
double t1, t2, t3, t4;
    t1 = det(x2 - x1, y2 - y1, x3 - x1, y3 - y1);
    t2 = det(x2 - x1, y2 - y1, x4 - x1, y4 - y1);
    t3 = det(x4 - x3, y4 - y3, x1 - x3, y1 - y3);
    t4 = det(x4 - x3, y4 - y3, x2 - x3, y2 - y3);
if(dcmp(t1) == 0 && inseg(x3, y3, x1, y1, x2, y2))
return 1;
if(dcmp(t2) == 0 && inseg(x4, y4, x1, y1, x2, y2))
return 1;
if(dcmp(t3) == 0 && inseg(x1, y1, x3, y3, x4, y4))
return 1;
if(dcmp(t4) == 0 && inseg(x2, y2, x3, y3, x4, y4))
return 1;
if(dcmp(t1) * dcmp(t2) < 0 && dcmp(t3) * dcmp(t4) < 0)
return 1;
return 0;
}
void calculate(int i, int j)
{
int p, q;
for(p = 0; p < vn[i]; p ++)
for(q = 0; q < vn[j]; q ++)
if(check(x[i][p], y[i][p], x[i][p + 1], y[i][p + 1], x[j][q], y[j][q], x[j][q + 1], y[j][q + 1]))
            {
                g[i][n[i] ++] = name[j];
                g[j][n[j] ++] = name[i];
return ;
            }
}
void printresult(int k)
{
int i, j;
if(n[k] == 0)
        printf("%c has no intersections\n", name[k]);
else if(n[k] == 1)
        printf("%c intersects with %c\n", name[k], g[k][0]);
else if(n[k] == 2)
        printf("%c intersects with %c and %c\n", name[k], g[k][0], g[k][1]);
else
    {
        printf("%c intersects with ", name[k]);
for(j = 0; j < n[k] - 1; j ++)
            printf("%c, ", g[k][j]);
        printf("and %c\n", g[k][n[k] - 1]);
    }
}
void solve()
{
int i, j, k, t;
    memset(n, 0, sizeof(n));
for(i = 0; i < N; i ++)
for(j = i + 1; j < N; j ++)
            calculate(i, j);
    memset(vis, 0, sizeof(vis));
for(i = 0; i < N; i ++)
    {
        t = 128;
for(j = 0; j < N; j ++)
if(!vis[j] && name[j] < t)
                k = j, t = name[j];
        vis[k] = 1;
        qsort(g[k], n[k], sizeof(g[k][0]), cmp);
        printresult(k);
    }
}
int main()
{
while(init())
    {
        solve();
        printf("\n");
    }
return 0;
}