题目来源:http://poj.org/problem?id=1022

题目大意:

  有一些4维的单位体积的立方体盒子,每个立方体有8个面。要用一个大的4为盒子将它们包起来,求最小的大盒子体积。

输入:第一行为测试用例数。每个用例的第一行为单位立方体数目n。接下来的n行每行为一个立方体的信息。每行第一个数字为还立方体的编号,接下来的8个整数分别为对应面相邻的立方体的编号。该面没有邻居则为0.(给出的都是单一刚体。)

输出:最小的能把这个由小4D立方体拼起来的形状的盒子的体积。


Sample Input

1
9
1 2 3 4 5 6 7 8 9
2 0 1 0 0 0 0 0 0
3 1 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0
5 0 0 1 0 0 0 0 0
6 0 0 0 0 0 1 0 0
7 0 0 0 0 1 0 0 0
8 0 0 0 0 0 0 0 1
9 0 0 0 0 0 0 1 0

Sample Output

81

本题题干描述得很复杂,想象起来也有一些抽象,其实很简单,跟3D的情况联系起来想就可以了。3D求包围盒的方法推广至4D即可。

 //////////////////////////////////////////////////////////////////////////
// POJ1022 Packing Unit 4D Cubes
// Memory: 300K Time: 16MS
// Language: C++ Result: Accepted
////////////////////////////////////////////////////////////////////////// #include <iostream>
#include <vector>
#include <map> using namespace std; class Cube {
public:
int x1p, x1n, x2p, x2n, x3p, x3n, x4p, x4n;
};
class Pos {
public:
int id;
int x1, x2, x3, x4;
}; int main() {
int ncase;
cin >> ncase;
for (int caseNo = ; caseNo <= ncase; ++caseNo) {
int n;
map<int, Cube> cubes;
cin >> n;
for (int i = ; i <= n; ++i) {
int id;
cin >> id;
Cube cube;
cin >> cube.x1p >> cube.x1n >> cube.x2p >> cube.x2n
>> cube.x3p >> cube.x3n >> cube.x4p >> cube.x4n;
cubes[id] = cube;
}
bool ok = true;
vector<Pos> solid;
Pos firstPos;
firstPos.id = (*cubes.begin()).first;
firstPos.x1 = firstPos.x2 = firstPos.x3 = firstPos.x4 = ;
solid.push_back(firstPos);
for (map<int, Cube>::iterator itc = cubes.begin(); itc != cubes.end(); ++itc) {
Cube cube1;
int id = (*itc).first;
int x1p = (*itc).second.x1p;
//x1p
if (x1p != ) {
if (cubes[x1p].x1n != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x1p;
pos.x1 = (*its).x1 + ;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x1p) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
} //x1n
int x1n = (*itc).second.x1n;
if (x1n != ) {
if (cubes[x1n].x1p != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x1n;
pos.x1 = (*its).x1 - ;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x1n) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
} //x2p
int x2p = (*itc).second.x2p;
if (x2p != ) {
if (cubes[x2p].x2n != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x2p;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2 + ;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x2p) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
}
//x2n
int x2n = (*itc).second.x2n;
if (x2n != ) {
if (cubes[x2n].x2p != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x2n;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2 - ;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x2n) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
} //x3p
int x3p = (*itc).second.x3p;
if (x3p != ) {
if (cubes[x3p].x3n != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x3p;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3 + ;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x3p) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
}
//x3n
int x3n = (*itc).second.x3n;;
if (x3n != ) {
if (cubes[x3n].x3p != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x3n;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3 - ;
pos.x4 = (*its).x4;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x3n) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
}
//x4p
int x4p = (*itc).second.x4p;
if (x4p != ) {
if (cubes[x4p].x4n != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x4p;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4 + ;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x4p) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
}
//x4n
int x4n = (*itc).second.x4n;
if (x4n != ) {
if (cubes[x4n].x4p != id) {
ok = false;
break;
}
bool f = true;
Pos pos;
for (vector<Pos>::iterator its = solid.begin(); its != solid.end(); ++its) {
if (f == false) break;
if ((*its).id == id) {
pos.id = x4n;
pos.x1 = (*its).x1;
pos.x2 = (*its).x2;
pos.x3 = (*its).x3;
pos.x4 = (*its).x4 - ;
for (vector<Pos>::iterator itr = solid.begin(); itr != solid.end(); ++itr) {
if ((*itr).id == x4n) {
f = false;
break;
}
}
}
}
if (f == true) {
solid.push_back(pos);
}
}
}
if (solid.size() != n) {
ok = false;
}
if (ok == false) {
cout << "Inconsistent" << endl;
continue;
}
int x1min = ;
int x1max = -;
int x2min = ;
int x2max = -;
int x3min = ;
int x3max = -;
int x4min = ;
int x4max = -;
for (vector<Pos>::iterator it = solid.begin(); it != solid.end(); ++it) {
if (x1min >(*it).x1) x1min = (*it).x1;
if (x1max < (*it).x1) x1max = (*it).x1;
if (x2min >(*it).x2) x2min = (*it).x2;
if (x2max < (*it).x2) x2max = (*it).x2;
if (x3min >(*it).x3) x3min = (*it).x3;
if (x3max < (*it).x3) x3max = (*it).x3;
if (x4min >(*it).x4) x4min = (*it).x4;
if (x4max < (*it).x4) x4max = (*it).x4;
}
int vol = (x1max - x1min + ) * (x2max - x2min + ) * (x3max - x3min + ) * (x4max - x4min + );
cout << vol << endl;
}
system("pause");
return ;
}

POJ1022 Packing Unit 4D Cubes的更多相关文章

  1. POJ题目排序的Java程序

    POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value ...

  2. Daily Query

    -- GI Report SELECT A.PLPKLNBR, D.DNDNHNBR, F.DNSAPCPO, C.PPPRODTE, A.GNUPDDTE GI_DATE, B.INHLDCDE, ...

  3. 【Moqui业务逻辑翻译系列】Shipment Receiver Receives Shipment with Packing Slip but no PO

    Shipment Receiver receives shipment. It has invoice tucked into it. Receiver records vendor name, ve ...

  4. ABP(现代ASP.NET样板开发框架)系列之12、ABP领域层——工作单元(Unit Of work)

    点这里进入ABP系列文章总目录 基于DDD的现代ASP.NET开发框架--ABP系列之12.ABP领域层——工作单元(Unit Of work) ABP是“ASP.NET Boilerplate Pr ...

  5. ABP源码分析十:Unit Of Work

    ABP以AOP的方式实现UnitOfWork功能.通过UnitOfWorkRegistrar将UnitOfWorkInterceptor在某个类被注册到IOCContainner的时候,一并添加到该类 ...

  6. Failed to stop iptables.service: Unit iptables.service not loaded.

    redhat 7 [root@lk0 ~]# service iptables stop Redirecting to /bin/systemctl stop iptables.service Fai ...

  7. 4D卓越团队-两天培训总结

    上周末参加了公司组织的领导力培训课程-4D卓越团队(创业型团队领导力训练项目),感觉有一些用,在这里分享一下. 课前游戏 培训老师先带我们做了一个游戏:每一个人,在同时参加培训的人中找到另外的 6 个 ...

  8. VS2012 Unit Test 个人学习汇总(含目录)

    首先,给出MSDN相关地址:http://msdn.microsoft.com/en-us/library/Microsoft.VisualStudio.TestTools.UnitTesting.a ...

  9. VS2012 Unit Test —— 我对IdleTest库动的大手术以及对Xml相关操作进行测试的方式

    [1]我的IdleTest源码地址:http://idletest.codeplex.com/ [2]IdleTest改动说明:2013年10月份在保持原有功能的情况下对其动了较大的手术,首先将基本的 ...

随机推荐

  1. Mule ESB学习【转-结合了网络上的一些资源】

    1.SOA标准之一:SCA架构思想 SOA在Java领域有两套标准:一个是SUN推出的JBI(没有得到BEA和IBM的承认),另外一个是:IBM和BEA等公司推出的SCA和SDO标准. JBI之关注J ...

  2. CTS 2019 Pearl

    CTS2019 Pearl 每种颜色有奇偶两种情况,只有偶数和只有奇数个,其对应的的指数型生成函数分别为 \(\frac{e^x+e^{-x}}2,\frac{e^x-e^{-x}}2\) 考虑选出现 ...

  3. BurpSuite工具应用及重放攻击实验

    一.BurpSuite工具介绍 BurpSuite是用于攻击web 应用程序的集成平台.它包含了许多工具,并为这些工具设计了许多接口,以促进加快攻击应用程序的过程.所有的工具都共享一个能处理并显示HT ...

  4. bzoj 2096 [POI2004]ZAW——二进制枚举

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2069 可以把直接相连的点分成  从1点出的一部分  和  走向1点的一部分.多起点最短路就和 ...

  5. 如何利用pyenv 和virtualenv 在单机上搭建多版本python 虚拟开发环境

    pyenv 和virtualenv分别是干什么的? pyenv帮助你在一台机上建立多个版本的python环境, 并提供方便的切换方法. virtualenv则就是将一个目录建立为一个虚拟的python ...

  6. 【转】 Pro Android学习笔记(四四):Dialog(1):触发Dialog

    目录(?)[-] 创建dialog fragment Activity显示对话框 Android提供alert.prompt.pick-list,单选.多选,progress.time-picker和 ...

  7. CRT Library Features

    CRT Library Features The new home for Visual Studio documentation is Visual Studio 2017 Documentatio ...

  8. cmake opencv,dlib 编译静态库 1

    无论windows,linux 所有的库 ,dlib,opencv 通过cmake-gui 设置好静态库, 动态库,和其他各个选项 Tips: cmake 优先级用cmake-gui,因为命令太多,容 ...

  9. Linux系统中‘dmesg’命令处理故障和收集系统信息的7种用法

    转自:https://linux.cn/article-3587-1.html 'dmesg'命令显示linux内核的环形缓冲区信息,我们可以从中获得诸如系统架构.cpu.挂载的硬件,RAM等多个运行 ...

  10. Linux编程里getopt_long_only函数用法详解

    在程序中难免需要使用命令行选项,可以选择自己解析命令行选项,但是有现成的,何必再造轮子.下面介绍使用getopt_long_only和getopt_long(两者用法差不多)解析命令行选项. 程序中主 ...