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. Maven发布项目丢失Mybatis Mapper包的映射问题

    由于一些eclipse版本问题,mybatis的mapper包中的sql文件没有被打进包,需要在pom中加入: <build> <!--配置打包时不过滤非java文件开始 --> ...

  2. ubuntu 上采用nginx做rtmp 直播 服务器

    首先安装必要的依赖库   sudo apt-get install autoconf automake sudo apt-get install libpcre3 libpcre3-dev   安装 ...

  3. 股神小D

    题目大意: 给定一棵树,每一条边有$L,R$两种权值,求有多少条路径满足$\max(L)\leq\min(R)$. 解法$1-$点分治$+$二维数点 统计树上的路径应首先想到点分治,我们很显然可以搜出 ...

  4. BZOJ5372: [Pkusc2018]神仙的游戏

    BZOJ5372: [Pkusc2018]神仙的游戏 https://lydsy.com/JudgeOnline/problem.php?id=5372 分析: 如果\(len\)为\(border\ ...

  5. BZOJ3033 太鼓达人

    3033: 太鼓达人 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 690  Solved: 497[Submit][Status][Discuss] ...

  6. codevs1060 搞笑世界杯

    题目描述 Description 随着世界杯小组赛的结束,法国,阿根廷等世界强队都纷纷被淘汰,让人心痛不已. 于是有 人组织了一场搞笑世界杯,将这些被淘汰的强队重新组织起来和世界杯一同比赛.你和你的朋 ...

  7. HDOJ1548(DFS超内存,BFS过了)

    DFS代码 #include<iostream> #include<cstdio> using namespace std; #define Min(a,b) (a<b) ...

  8. 使用TRY CATCH进行SQL Server异常处理

    TRY...CATCH是Sql Server 2005/2008令人印象深刻的新特性.提高了开发人员异常处理能力.没有理由不尝试一下Try.. Catch功能. *      TRY 块 - 包含可能 ...

  9. 利用Ssocks访问国外网站(yutube/google等)

    ***开源项目:https://github.com/***/ 本例使用的是针对windows系统的c-sharp版本:https://github.com/***/***-windows 运行*** ...

  10. Spring 框架学习整理

    JDBC操作数据库的基本入门中存在什么问题? *   导致驱动注册两次是个问题,但不是严重的. *   严重的问题:是当前类和mysql的驱动类有很强的依赖关系. *      当我们没有驱动类的时候 ...