UVA Solution 184 - Laser Lines - Solution in C++ | Volume 1 - Online Judge Solution

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Sunday, April 30, 2017

UVA Solution 184 - Laser Lines - Solution in C++ | Volume 1

UVA Solution 184 - Laser Lines - Solution in C++ | Volume 1


UVA Online Judge Solution 184 - Laser Lines | Volume 1


Problem Name: Laser Lines
Problem Number : UVA - 184
Online Judge : UVA Online Judge Solution
Volume: 1
Solution Language : C plus plus

UVA Solution 184 Code in CPP:


#include <stdio.h>
#include <stdlib.h>
typedef struct {
 int x, y, i;
}Point;

int cmp(const void *i, const void *j) {
 Point *x, *y;
 x = (Point *)i, y = (Point *)j;
 if(x->x != y->x)
  return x->x - y->x;
    if(x->y != y->y)
        return x->y - y->y;
    return x->i - y->i;
}
int main() {
    Point D[301];
    int x, y, i, j, k, l;
    while(scanf("%d %d", &x, &y) == 2) {
        if(!x && !y)
            break;
        D[0].x = x, D[0].y = y;
        int n = 1;
        while(scanf("%d %d", &x, &y) == 2) {
            if(!x && !y)
                break;
            D[n].x = x, D[n].y = y, n++;
        }
        qsort(D, n, sizeof(Point), cmp);
        for(i = 0; i < n; i++)
            D[i].i = i;
        int map[301][301] = {};
        int flag = 0;
        for(i = 0; i < n; i++) {
            for(j = i+1; j < n; j++) {
                if(map[i][j] == 0) {
                    map[i][j] = 1;
                    int ans[301], aidx = 2;
                    ans[0] = i, ans[1] = j;
                    for(k = j+1; k < n; k++) {
                        if((D[i].y-D[j].y)*(D[j].x-D[k].x) == (D[i].x-D[j].x)*(D[j].y-D[k].y)) {
                            ans[aidx++] = k;
                        }
                    }
                    if(aidx >= 3) {
                        if(!flag)
                            puts("The following lines were found:");
                        flag = 1;
                        for(k = 0; k < aidx; k++)
                            for(l = 0; l < aidx; l++)
                                map[ans[k]][ans[l]] = 1;
                        for(k = 0; k < aidx; k++)
                            printf("(%4d,%4d)", D[ans[k]].x, D[ans[k]].y);
                        puts("");
                    }
                }
            }
        }
        if(!flag)
            puts("No lines were found");
    }
    return 0;
}

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