本文主要是介绍POJ 3668 Game of Lines/洛谷 P2665 水题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目背景
Farmer John最近发明了一个游戏，来考验自命不凡的贝茜。
题目描述
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.
游戏开始的时 候，FJ会给贝茜一块画着N (2 <= N <= 200)个不重合的点的木板，其中第i个点 的横、纵坐标分别为X_i和Y_i (-1,000 <= X_i <=1,000； -1,000 <= Y_i <= 1,000)。 贝茜可以选两个点画一条过它们的直线，当且仅当平面上不存在与画出直线 平行的直线。游戏结束时贝茜的得分，就是她画出的直线的总条数。为了在游戏 中胜出，贝茜找到了你，希望你帮她计算一下最大可能得分。
输入输出格式
输入格式：第1行: 输入1个正整数：N
第2..N+1行: 第i+1行用2个用空格隔开的整数X_i、Y_i，描述了点i的坐标
输出格式：第1行: 输出1个整数，表示贝茜的最大得分，即她能画出的互不平行的线段数
输入输出样例
4 -1 1 -2 0 0 0 1 1
4
说明
贝茜能画出以下4种斜率的直线：-1，0，1/3以及1。
USACO2008 Feb T1
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
#define N 40005
int n;
double a[205],b[205],c[N];
//c保存斜率
int main()
{ while(scanf("%d",&n)!=EOF) { for(int i=1;i<=n;i++)scanf("%lf%lf",&a[i],&b[i]);int sum=0;for(int i=1;i<n;i++)for(int j=i+1;j<=n;j++){if(a[i]==a[j]) c[++sum]=0;else if(b[i]==b[j]) c[++sum]=N;else c[++sum]=(a[i]-a[j])/(b[i]-b[j]) ;}sort(c+1,c+sum+1);int ans=0;for(int i=1;i<=sum;i++){if(c[i]-c[i-1]>=1e-10) ans++; //注意double类型判断是否为0}printf("%d\n",ans+1);} return 0; } 这篇关于POJ 3668 Game of Lines/洛谷 P2665 水题的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
