Game of Lines

Time Limit: 1000MS Memory Limit: 65536K

Description

Farmer John has challenged Bessie to the following game: FJ has a board with dots marked at N (2 ≤ N ≤ 200) distinct lattice points. Dot i has the integer coordinates Xi and Yi (-1,000 ≤ Xi ≤ 1,000; -1,000 ≤ Yi ≤ 1,000).

Bessie can score a point in the game by picking two of the dots and drawing a straight line between them; however, she is not allowed to draw a line if she has already drawn another line that is parallel to that line. Bessie would like to know her chances of winning, so she has asked you to help find the maximum score she can obtain.

Input

  • Line 1: A single integer: N
  • Lines 2…N+1: Line i+1 describes lattice point i with two space-separated integers: Xi and Yi.

Output

  • Line 1: A single integer representing the maximal number of lines Bessie can draw, no two of which are parallel.

Sample Input

4
-1 1
-2 0
0 0
1 1

Sample Output

4

第一种方法：

思路：

水题，就是会用模板库就行了，每次都存进斜率，因为用的set所以最后直接输出size就行了。

#include <iostream>
#include <set>
#include <cstdio>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 210;
int main() {
    ios::sync_with_stdio;
    int n;
    set<double> st;
    double x[maxn], y[maxn];
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%lf %lf", &x[i], &y[i]);
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            double k;
            if (x[i] == x[j]) k = inf;
            else k = (y[i] - y[j]) / (x[i] - x[j]);
            st.insert(k);
        }
    }
    printf("%d", st.size());
    return 0;
}

第二种方法：

思路：

先将斜率排序，然后再进行比较。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 210;
int main() {
    ios::sync_with_stdio;
    int n;
    vector<double> v;
    double x[maxn], y[maxn];
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%lf %lf", &x[i], &y[i]);
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            double k;
            if (x[i] == x[j]) k = inf;
            else k = (y[i] - y[j]) / (x[i] - x[j]);
            v.push_back(k);
        }
    }
    int sum = 0;
    sort(v.begin(), v.end());
    for (int i = 1; i < v.size(); i++) {
        if (v[i] != v[i - 1]) sum++;
    }
    printf("%d", sum + 1);
    return 0;
}