Codeforces_733D
3 seconds
256 megabytes
standard input
standard output
Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya's idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere.
Zahar has n stones which are rectangular parallelepipeds. The edges sizes of the i-th of them are ai, bi and ci. He can take no more than two stones and present them to Kostya.
If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way.
Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.
The first line contains the integer n (1 ≤ n ≤ 105).
n lines follow, in the i-th of which there are three integers ai, bi and ci (1 ≤ ai, bi, ci ≤ 109) — the lengths of edges of the i-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.
In the first line print k (1 ≤ k ≤ 2) the number of stones which Zahar has chosen. In the second line print k distinct integers from 1 to n — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to n in the order as they are given in the input data.
You can print the stones in arbitrary order. If there are several answers print any of them.
6
5 5 5
3 2 4
1 4 1
2 1 3
3 2 4
3 3 4
1
1
7
10 7 8
5 10 3
4 2 6
5 5 5
10 2 8
4 2 1
7 7 7
2
1 5
In the first example we can connect the pairs of stones:
- 2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
- 2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5 respectively.
- 2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5
- 4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
- 5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5
Or take only one stone:
- 1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5
- 2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
- 3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5
- 4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5
- 5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
- 6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5
It is most profitable to take only the first stone.
题意:给定n个长方体的长宽高,一次可以取一个,或者如果两个长方体有相同面,那么可以将这两个长方体通过相同面连接起来,最多可以连两个长方体。问要取的长方体的内接球的半径最大,应该取哪一个或者哪两个。
思路:先算出取一个能取的最大值,再算两个的。长方体内接球的半径受限于最短边,也就是说如果要将两个拼接起来,若不增加最短边,那么结果不可能大于取一个的最大值。接下来就暴力了。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
using namespace std;
#define N 100005
#define INF 1e9+5
struct Node
{
int a,b,c,index;
} node[N]; bool cmp(Node x,Node y)
{
if(x.a>y.a)
return ;
else if(x.a==y.a)
{
if(x.b>y.b)
return ;
else if(x.b==y.b)
{
if(x.c>y.c)
return ;
else
return ;
}
else
return ;
}
else
return ;
} int getR(int a,int b,int c)
{
int minx=INF;
minx=min(minx,a);
minx=min(minx,b);
minx=min(minx,c);
return minx;
} int main()
{
int n;
scanf("%d",&n);
int num[];
for(int i=; i<n; i++)
{
scanf("%d%d%d",&num[],&num[],&num[]);
sort(num,num+);
node[i].a=num[];
node[i].b=num[];
node[i].c=num[];
node[i].index=i+;
}
sort(node,node+n,cmp);
int aa=-,bb=-,maxn=;
for(int i=; i<n; i++)
{
int tmp=getR(node[i].a,node[i].b,node[i].c);
if(tmp>maxn)
{
maxn=tmp;
aa=node[i].index;
}
}
for(int i=; i<n; i++)
{
if(node[i].a==node[i-].a&&node[i].b==node[i-].b)
{
int tmp=getR(node[i].a,node[i].b,node[i].c+node[i-].c);
if(tmp>maxn)
{
maxn=tmp;
aa=node[i].index;
bb=node[i-].index;
}
}
}
if(aa>&&bb>)
{
printf("2\n");
printf("%d %d\n",aa,bb);
}
else
{
printf("1\n");
printf("%d",aa);
}
return ;
}
Codeforces_733D的更多相关文章
随机推荐
- Linux运行级别研究(转)
Linux系统中的运行级别 7种运行级别 运行级别(Runlevel)指的是Unix或者Linux等类Unix操作系统的运行模式,不同的运行模式下系统的功能也有所有不同.Linux 系统下通常分为7种 ...
- [Vue-rx] Watch Vue.js v-models as Observable with $watchAsObservable and RxJS
You most likely already have data or properties in your template which are controlled by third-party ...
- 【Cocos2dx游戏开发】Cocos2d-x简介
一.简介 最近在做一个Android下的卡牌游戏--<九州幻想>开发项目,而我们使用的引擎是Cocos2dx,所以想要写写笔记来记录一下项目中的收获.当然首先稍微介绍一下Cocos2d-x ...
- Wireshark 抓包遇到 you don’t have permission to capture on that device mac 错误的解决方案
Wireshark 抓包遇到 you don’t have permission to capture on that device mac 错误的解决方案 上次有篇博客讲了如何利用wireshark ...
- 深入struts2(二) ---stuts2长处和主要包、类功能
1.1 Struts2 上节已讲.struts2在webwork基础发展起来的mvc框架.MVC框架相信一般码农都比較了解,这里不再重说. 在这里只对于一下struts1,struts2做了哪 ...
- Angular常用标记
(如果没有特别指明,则所有的HTML元素都支持该标记) (如果没有特别指明,则 AngularJS 指令不会覆盖原生js的指令) 1.数据绑定类: 1.插值语法:{{}} 2.标签内容绑定:ng-bi ...
- 一个APP爆炸的时代,假设没有wifi
我们每天都离不开的微信,又传来了一个新消息.你造?微信公众平台新增了设备功能.眼下可支持可穿戴设备,将来呢,前景不可限量!能够想象,"后果"是我们越来越离不开微信,依附于它.这样的 ...
- BeautifulSoup中各种html解析器的比較及使用
Beautiful Soup解析器比較 ·Beautiful Soup支持各种html解析器.包含python自带的标准库.还有其它的很多第三方库模块. 当中一个就是lxml parser,至于lxm ...
- android Activity初次的启动的时候播放声音
代码例如以下: private MediaPlayer mMediaPlayer; mMediaPlayer = new MediaPlayer(); mMediaPlayer = MediaPlay ...
- Batch 拷贝远程机器文件到本机指定目录下
net use * /del /yesNET USE Y: \\远程机IP\d$ 登录密码 /user:domain\登录用户 set sourcePath="Y:\DOAutomatio ...