c++. placing points on straight lines - c++

Question

In one of the largest cities in Bytland, Bytesburg, construction of the second stage of the metro is underway. During the construction of the first stage, N stations were built that were not interconnected. According to the master plan, the metro in Bytesburg should consist of no more than two lines. each metro line is straight. The president of the company responsible for laying the lines wants to make sure that no more than two metro lines can be laid, so that all stations built lie on at least one of the two
Exaple 1
input:
6
0 1
1 1
2 1
0 2
1 3
2 2
output:
no
Example 2
input:
6
2 2
4 6
1 0
2 1
6 1
1 1
output:
yes
I wrote the code, but on the seventh test on the testing system, it gives the wrong answer. Help me please. This is my code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct line {
int k, b;
bool x = false;
long long c = 0;
};
int n;
vector<line> lines;
vector<pair<int, int>> p;
bool Is3PointsOnLine(pair<float, float> p1, pair<float, float> p2, pair<float, float> p3) {
return ((p3.first - p1.first)*(p2.second - p1.second) == (p3.second - p1.second)*(p2.first - p1.first));
}
bool PointIsOnLine(line l, int x, int y) {
if (!l.x) {
if (y == ((l.k * x) + l.b))
return true;
}
else {
if (x == l.b)
return true;
}
return false;
}
line f(pair<int, int> c1, pair<int, int> c2) {
int x1 = c1.first;
int y1 = c1.second;
int x2 = c2.first;
int y2 = c2.second;
line ans;
if (x2 != x1) {
ans.k = (y2 - y1) / (x2 - x1);//fix
ans.b = -(x1 * y2 - x1 * y1 - x2 * y1 + x1 * y1) / (x2 - x1);
ans.x = false;
ans.c = 0;
}
else {
ans.k = 0;
ans.b = x1;
ans.x = true;
ans.c = 0;
}
return ans;
}
int a() {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (j != i) {
for (int k = 0; k < n; k++) {
if ((k != i) && (k != j) && (Is3PointsOnLine(p[i], p[j], p[k]))) {
lines.push_back(f(p[i], p[j]));
return 0;
}
}
}
}
}
}
int main() {
cin >> n;
p.resize(n);
vector<int> not_used;
for (int i = 0; i < n; i++)
cin >> p[i].first >> p[i].second;
if (n < 5) {
cout << "yes";
return 0;
}
else {
a();
if (lines.size() == 0) {
cout << "no";
return 0;
}
pair<int, int> e = { -8,-8 };
for (int i = 0; i < n; i++) {
if (PointIsOnLine(lines[0], p[i].first, p[i].second)) {
lines[0].c++;
}
else if (e.first == -8) {
e.first = i;
}
else if (e.second == -8) {
e.second = i;
lines.push_back(f(p[e.first], p[e.second]));
for (int j = 0; j <= i; j++) {
if (PointIsOnLine(lines[1], p[j].first, p[j].second)) {
lines[1].c++;
}
}
e = { -5,-5 };
}
else if (PointIsOnLine(lines[1], p[i].first, p[i].second)) {
lines[1].c++;
}
else {
cout << "no";
return 0;
}
}
if (lines[0].c+1 >= n && e.first != -2 && e.second == -8) {
cout << "yes";
return 0;
}
else if (lines[0].c + lines[1].c >= n) {
cout << "yes";
return 0;
}
else {
cout << "no";
return 0;
}
}
return 0;
}
Code v2
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct line {
pair<int, int> p1, p2;
long long c = 0;
};
int n;
vector<line> lines;
vector<pair<int, int>> p;
bool Is3PointsOnLine(pair<float, float> p1, pair<float, float> p2, pair<float, float> p3) {
return ((p3.first - p1.first)*(p2.second - p1.second) == (p3.second - p1.second)*(p2.first - p1.first));
}
//лежит ли точка на прямой
bool PointIsOnLine(line l, int x, int y) {
return Is3PointsOnLine(l.p1, l.p2, {x , y});
}
int a() {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (j != i) {
for (int k = 0; k < n; k++) {
if ((k != i) && (k != j) && (Is3PointsOnLine(p[i], p[j], p[k]))) {
line sdafsadf;
sdafsadf.p1 = p[i]; sdafsadf.p2 = p[j];
lines.push_back(sdafsadf);
return 0;
}
}
}
}
}
}
int main() {
cin >> n;
p.resize(n);
vector<int> not_used;
for (int i = 0; i < n; i++)
cin >> p[i].first >> p[i].second;
if (n < 5) {
cout << "yes";
return 0;
}
else {
//ищем первую прямую
a();
if (lines.size() == 0) {
cout << "no";
return 0;
}
pair<int, int> e = { -8,-8 };
for (int i = 0; i < n; i++) {
if (PointIsOnLine(lines[0], p[i].first, p[i].second)) {
lines[0].c++;
}
else if (e.first == -8) {
e.first = i;
}
else if (e.second == -8) {
e.second = i;
line sdafsadf;
sdafsadf.p1 = p[e.first]; sdafsadf.p2 = p[e.second];
lines.push_back(sdafsadf);
for (int j = 0; j <= i; j++) {
if (PointIsOnLine(lines[1], p[j].first, p[j].second)) {
lines[1].c++;
}
}
e = { -5,-5 };
}
else if (PointIsOnLine(lines[1], p[i].first, p[i].second)) {
lines[1].c++;
}
else {
cout << "no";
return 0;
}
}
if (lines[0].c+1 >= n && e.first != -2 && e.second == -8) {
cout << "yes";
return 0;
}
else if ((lines[0].c + lines[1].c >= n) && (lines[0].p1 != lines[1].p1) && (lines[0].p1 != lines[1].p2) && (lines[0].p2 != lines[1].p1) && (lines[0].p2 != lines[1].p2)) {
cout << "yes";
return 0;
}
else {
cout << "no";
return 0;
}
}
return 0;
}

Answer 1

Your function f is horribly broken because it uses integer division. Then everything after that which relies on k and b, including PointIsOnLine, will give wrong results. Is3PointsOnLine seems ok, change your line struct to store two points, not slope+intercept, and then PointInOnLine can just call Is3PointsOnLine.
I'm amazed that you passed any non-trivial test cases with such a bug.
This also will not work
if (lines[0].c + lines[1].c >= n)
because you aren't testing that lines[1] isn't a duplicate of lines[0].
Also, you can reach lines[0].c + lines[1].c == n and still miss one station completely, if another station lies at the intersection of the lines and gets double-counted.