题目:

Go Northwest! is a game usually played in the park main hall when occasional rainy weather discourages the visitors from enjoying outdoor attractions.

The game is played by a pair of players and it is based entirely on luck, the players can hardly influence its outcome.

The game plan consists of one large map on the wall with N distinct towns displayed on it. Before the start of the game, the leader of the game hands a bowl filled with N identical balls to each player. Inside each ball, there is a reference to some town on the map. Each bowl contains references to all towns on the map.

When the game starts, one of the players randomly draws one ball from his/her bowl, opens it and reveals the town inside. Next, the other player randomly draws one ball form his/her bowl, opens it and also reveals the town inside.

If both towns are the same, the game ends in a draw. If the towns are not the same, the leader of the game highlights the towns on the map. If one of the towns lies exactly Northwest to the other, the first player wins the game. If one of the towns lies exactly Northeast to the other, the second player wins the game. In all other cases, the game ends in a draw. Two towns are considered to lie exactly Northwest or Northeast from each other, if the angle between the line connecting the towns and the horizontal axis is exactly 45 degrees.

It is clear that the chance of winning the game depends substantially on the layout of the towns on the map. All town coordinates are known in advance. You are asked to find the probability that there is a winner in the game.

Input Specification:

There are more test cases. Each case starts with a line containing one integer N (1 ≤ N ≤ 105 ) specifying the number of towns on the map. Next, there are N lines each of which contains two integers x and y separated by space and specifying the coordinates of one town on the map. The coordinates are standard Cartesian coordinates in the plane. The absolute value of any coordinate does not exceed 109 .

Output Specification:

For each test case, print a single line with one floating point number P denoting the probability that there is a winner in the game. P should be printed with the maximum allowed error of 10-6.

题意:

给出一些点的坐标,两个玩家随机选取两个点(这两个点可以是同一个),如果这两个点的斜率是1或者是-1,则这两个玩家就能分出胜负,否则游戏结束。问给出的点中,所有的选择中有玩家胜出的概率是多少。

思路:

将给出的每一个点,分别以1和-1的斜率做直线与y轴交于两个点,将这两个点的y坐标的大小记录下来。那么遍历这个map容器,答案就是对映射值(k)的k*(k-1)求和,最终结果除以n*n。

PS:

一定要注意数据范围!!!!

代码:

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2e5+; int main() {
ll n;
while(scanf("%lld",&n)!=EOF) {
map<ll, ll> add;
//map<int, int> sub;
for(ll i = ; i<n; i++){
ll x,y;
scanf("%lld%lld",&x,&y);
add[y+x]++;
add[y-x]++;
}
map<ll, ll>::iterator it;
ll sum = ,temp = n*n;
for(it = add.begin(); it!=add.end(); it++){
sum += it->second*(it->second-);
}
printf("%.6f\n",1.0*sum/temp);
}
return ;
}
/*
PutIn:
4
1 2
2 1
1 1
2 2
3
3 1
2 2
1 3
PutOut:
0.25
0.666667
*/

Gym - 101670H Go Northwest!(CTU Open Contest 2017 思维题+map)的更多相关文章

  1. Gym - 101670A Amusement Anticipation(CTU Open Contest 2017 签到题)

    题目&题意: 倒着找处于最后位置的等差数列的开头的位置. 例: 1 5 3 4 5 6 3 4 5 6是等差数列,它的开头的位置是3 PS: 读题真的很重要!!!!多组输入,上来就读错了!! ...

  2. (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest(爽题)

    layout: post title: (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest(爽题) author: " ...

  3. Gym - 101670H Dark Ride with Monsters(CTU Open Contest 2017 贪心)

    题目: A narrow gauge train drives the visitors through the sequence of chambers in the Dark Ride attra ...

  4. Gym - 101670F Shooting Gallery(CTU Open Contest 2017 区间dp)

    题目&题意:(有点难读...) 给出一个数字序列,找出一个区间,当删除这个区间中的两个相同的数字后,只保留这两个数字之间的序列,然后继续删除相同的数字,问最多可以实行多少次删除操作. 例如: ...

  5. Gym - 101670G Ice cream samples(CTU Open Contest 2017 尺取法)

    题目: To encourage visitors active movement among the attractions, a circular path with ice cream stan ...

  6. Gym - 101670E Forest Picture (CTU Open Contest 2017 模拟)

    题目: https://cn.vjudge.net/problem/1451310/origin 题意&思路: 纯粹模拟. 大体题意是这样的: 1.有人要在一个10-9<=x<=1 ...

  7. Gym - 101670C Chessboard Dancing(CTU Open Contest 2017 找规律)

    题目:链接 思路: 多画出几个情况就可以找出规律来了 Knight (当大于2的时候只要两种颜色相间出现就可以了) King(当大于等于3的时候,总可以用四种形式来补色,具体如下)  Bishop(斜 ...

  8. Gym - 101670B Pond Cascade(CTU Open Contest 2017 贪心,二分)

    题目: The cascade of water slides has been installed in the park recently and it has to be tested. The ...

  9. Gym - 101670J Punching Power(CTU Open Contest 2017 最大独立集)

    题目: The park management finally decided to install some popular boxing machines at various strategic ...

随机推荐

  1. 01排序的Java实现

    比赛描写叙述: 的个数同样时再按ASCII码值排序. 输入: 个字符. 输出: 串的顺序.使得串按基本描写叙述的方式排序. 例子输入: 例子输出: 被AC的代码例如以下: import java.ut ...

  2. Ubuntu 安装Consolas字体

    下载 YaHei.Consolas https://code.google.com/p/uigroupcode/downloads/list 安装 1.创建文件夹 sudo mkdir -p /usr ...

  3. [Codeforces 460C] Present

    [题目链接] https://codeforces.com/contest/460/problem/C [算法] 二分 + 贪心 要求最小值最大 , 我们不妨二分最小值 , 若一盆花的高度小于二分的值 ...

  4. 深入理解JMM(Java内存模型) --(七)总结

    JMM 掌管着一个线程对内存的动作 (读和写)影响其他线程对内存的动作的方式.由于使用处理器寄存器和预处理 cache 来提高内存访问速度带来的性能提升,Java 语言规范(JLS)允许一些内存操作并 ...

  5. Face alignment at 3000FPS via Regressing Local Binrary features 理解

    这篇是Ren Shaoqing发表在cvpr2014上的paper,论文是在CPR框架下做的,想了解CPR的同学可以参见我之前的博客,网上有同学给出了code,该code部分实现了LBF,链接为htt ...

  6. bzoj 1631: [Usaco2007 Feb]Cow Party【spfa】

    正反加边分别跑spfa最短路,把两次最短路的和求个max就是答案 #include<iostream> #include<cstdio> #include<queue&g ...

  7. POJ 3608 旋转卡壳

    思路: 旋转卡壳应用 注意点&边  边&边  点&点 三种情况 //By SiriusRen #include <cmath> #include <cstdi ...

  8. 转 js实践篇:例外处理Try{}catch(e){}

    程序开发中,编程人员经常要面对的是如何编写代码来响应错误事件的发生,即例外处理(exception handlers).如果例外处理代码设计得周全,那么最终呈现给用户的就将是一个友好的界面.否则,就会 ...

  9. pyinstaller遇到的坑

    最近接了一个python的活,具体的就不展开,大概就是需要搭建一个服务器,接收客户端上传文件,调用算法模型,然后返回相应的数据.算法模块用的是tensorflow模块,里面一大堆东西,网上看了很多,最 ...

  10. java 装饰者类

    装饰者模式:增强一个类的功能还可以让装饰者类之间互相装饰. 装饰者模式和继承的区别: 继承实现的增强类: 优点:代码结构清晰,而且实现简单 缺点:对于每一个的需要增强的类都要创建具体的子类来帮助其增强 ...