LightOJ - 1410 - Consistent Verdicts(规律)
链接:
https://vjudge.net/problem/LightOJ-1410
题意:
In a 2D plane N persons are standing and each of them has a gun in his hand. The plane is so big that the persons can be considered as points and their locations are given as Cartesian coordinates. Each of the N persons fire the gun in his hand exactly once and no two of them fire at the same or similar time (the sound of two gun shots are never heard at the same time by anyone so no sound is missed due to concurrency). The hearing ability of all these persons is exactly same. That means if one person can hear a sound at distance R1, so can every other person and if one person cannot hear a sound at distance R2 the other N-1 persons cannot hear a sound at distance R2 as well.
The N persons are numbered from 1 to N. After all the guns are fired, all of them are asked how many gun shots they have heard (not including their own shot) and they give their verdict. It is not possible for you to determine whether their verdicts are true but it is possible for you to judge if their verdicts are consistent. For example, look at the figure above. There are five persons and their coordinates are (1, 2), (3, 1), (5, 1), (6, 3) and (1, 5) and they are numbered as 1, 2, 3, 4 and 5 respectively. After all five of them have shot their guns, you ask them how many shots each of them have heard. Now if there response is 1, 1, 1, 2 and 1 respectively then you can represent it as (1, 1, 1, 2, 1). But this is an inconsistent verdict because if person 4 hears 2 shots then he must have heard the shot fired by person 2, then obviously person 2 must have heard the shot fired by person 1, 3 and 4 (person 1 and 3 are nearer to person 2 than person 4). But their opinions show that Person 2 says that he has heard only 1 shot. On the other hand (1, 2, 2, 1, 0) is a consistent verdict for this scenario so is (2, 2, 2, 1, 1). In this scenario (5, 5, 5, 4, 4) is not a consistent verdict because a person can hear at most 4 shots.
Given the locations of N persons, your job is to find the total number of different consistent verdicts for that scenario. Two verdicts are different if opinion of at least one person is different.
思路:
计算任意两点距离,不同种类数就是距离数
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
const int INF = 1e9;
const int MAXN = 710;
const int MOD = 1e9+7;
int x[MAXN], y[MAXN];
int n;
int len[MAXN*MAXN];
int GetLen(int i, int j)
{
return (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
}
int main()
{
int t, cnt = 0;
scanf("%d", &t);
while(t--)
{
printf("Case %d:", ++cnt);
scanf("%d", &n);
for (int i = 1;i <= n;i++)
scanf("%d%d", &x[i], &y[i]);
int pos = 0;
for (int i = 1;i <= n;i++)
{
for (int j = i+1;j <= n;j++)
len[++pos] = GetLen(i, j);
}
sort(len+1, len+1+pos);
int res = unique(len+1, len+1+pos)-(len+1);
printf(" %d\n", res+1);
}
return 0;
}
LightOJ - 1410 - Consistent Verdicts(规律)的更多相关文章
- 1410 - Consistent Verdicts(规律)
1410 - Consistent Verdicts PDF (English) Statistics Forum Time Limit: 5 second(s) Memory Limit: 32 ...
- LightOJ 1410 Consistent Verdicts(找规律)
题目链接:https://vjudge.net/contest/28079#problem/Q 题目大意:题目描述很长很吓人,大概的意思就是有n个坐标代表n个人的位置,每个人听力都是一样的,每人发出一 ...
- Fibsieve`s Fantabulous Birthday LightOJ - 1008(找规律。。)
Description 某只同学在生日宴上得到了一个N×N玻璃棋盘,每个单元格都有灯.每一秒钟棋盘会有一个单元格被点亮然后熄灭.棋盘中的单元格将以图中所示的顺序点亮.每个单元格上标记的是它在第几秒被点 ...
- Harmonic Number (II) LightOJ - 1245 (找规律?。。。)
题意: 求前n项的n/i 的和 只取整数部分 暴力肯定超时...然后 ...现在的人真聪明...我真蠢 觉得还是别人的题意比较清晰 比如n=100的话,i=4时n/i等于25,i=5时n/i等于20 ...
- Trailing Zeroes (III) LightOJ - 1138 不找规律-理智推断-二分
其实有几个尾零代表10的几次方但是10=2*510^n=2^n*5^n2增长的远比5快,所以只用考虑N!中有几个5就行了 代码看别人的: https://blog.csdn.net/qq_422797 ...
- Trailing Zeroes (III) LightOJ - 1138 二分+找规律
Time Limit: 2 second(s) Memory Limit: 32 MB You task is to find minimal natural number N, so that N! ...
- lightoj--1410--Consistent Verdicts(技巧)
Consistent Verdicts Time Limit: 5000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu Su ...
- 初次使用SQL调优建议工具--SQL Tuning Advisor
在10g中,Oracle推出了自己的SQL优化辅助工具: SQL优化器(SQL Tuning Advisor :STA),它是新的DBMS_SQLTUNE包. 使用STA一定要保证优化器是CBO模式下 ...
- LightOj 1245 --- Harmonic Number (II)找规律
题目链接:http://lightoj.com/volume_showproblem.php?problem=1245 题意就是求 n/i (1<=i<=n) 的取整的和这就是到找规律的题 ...
随机推荐
- MySQL中主键id不连贯重置处理办法
MySQL中有时候会出现主键字段不连续,或者顺序乱了,想重置从1开始自增,下面处理方法 先删除原有主键,再新增新主键字段就好了 #删除原有自增主键 ALTER TABLE appraiser_info ...
- Java开发笔记(一百一十四)利用Socket传输文本消息
前面介绍了HTTP协议的网络通信,包括接口调用.文件下载和文件上传,这些功能固然已经覆盖了常见的联网操作,可是HTTP协议拥有专门的通信规则,这些规则一方面有利于维持正常的数据交互,另一方面不可避免地 ...
- c# 用XmlWriter写xml序列化
using System.Text; using System.Xml; using System.Xml.Schema; using System.Xml.Serialization; using ...
- mysql中数据表记录的增删查改(1)
数据记录的增删改查 insert into `数据表名称` (`字段名称`, ...) values ('1', ...); delete from `数据表名称` where 子句; update ...
- NLP自然语言处理的开发环境搭建
NLP的开发环境搭建主要分为以下几步: Python安装 NLTK系统安装 Python3.5下载安装 下载链接:https://www.python.org/downloads/release/py ...
- vue中$router与$route的区别
$.router是VueRouter的实例,相当于一个全局的路由器对象.包含很多属性和子对象,例如history对象 $.route表示当前正在跳转的路由对象.可以通过$.route获取到name,p ...
- (二)pdf的构成综述
引自:https://blog.csdn.net/steve_cui/article/details/81948486 一个pdf文件主要是由4部分构成:文件头.文件体.交叉引用表.文件尾 文件头:用 ...
- 玩转Spring全家桶笔记 04 Spring的事务抽象、事务传播特性、编程式事务、申明式事务
1.Spring 的事务抽象 Spring提供了一致的事务模型 JDBC/Hibernate/Mybatis 操作数据 DataSource/JTA 事务 2.事务抽象的核心接口 PlatformTr ...
- CSS3 @font-face 规则
指定名为"myFirstFont"的字体,并指定在哪里可以找到它的URL: @font-face { font-family: myFirstFont; src: url('San ...
- native function 'Window_sendPlatformMessage' (4 arguments) cannot be found
https://github.com/pauldemarco/flutter_blue/issues/140 https://github.com/flutter/flutter/issues/168 ...