POJ1022 Packing Unit 4D Cubes
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
- 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 ...
- Daily Query
-- GI Report SELECT A.PLPKLNBR, D.DNDNHNBR, F.DNSAPCPO, C.PPPRODTE, A.GNUPDDTE GI_DATE, B.INHLDCDE, ...
- 【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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 优秀开源项目之一:视频监控系统iSpy
iSpy是一个开源的视频监控软件,目前已经支持中文.自己用了一下,感觉还是很好用的.翻译了一下它的介绍. iSpy将PC变成一个完整的安全和监控系统 iSpy使用您的摄像头和麦克风来检测和记录声音或运 ...
- 深入V8引擎-枚举+位运算实现参数配置
不知不觉都快月底了,看了看上一篇还是6号写的,惭愧惭愧,说好的坚持.为了证明没有偷懒(其实还是沉迷了一会dota2),先上一个图自证清白. 基本上从初始化引擎,到Isolate.handleScope ...
- bzoj 4823 & 洛谷 P3756 老C的方块 —— 最小割
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=4823 https://www.luogu.org/problemnew/show/P3756 ...
- js浮点数运算出现误差解决方案
1.数据展示类(使用 toPrecision 凑整并 parseFloat 转成数字后再显示) parseFloat(1.4000000000000001.toPrecision(12)) === 1 ...
- Python-Redis的发布与订阅
封装的redis_config # -*- coding:utf-8 -*- __author__ = "MuT6 Sch01aR" import redis class Redi ...
- linux命令-vim命令模式
编辑模式下 按 “esc” 进去命令模式 :wq 保存并退出 :w 保存 已写入 :q!不保存强制退出 :wq! 强制保存退出 :set nu 显示行号 :set nonu 去掉行号 :nohl 取 ...
- VMware虚拟机重置密码
1.给vmware虚拟机添加启动延时 1.1 编辑VMware的配置文件.vmx,开机就自动进入BIOS 加入一行:bios.forceSetupOnce = " ...
- MySQL的limit优化
mysql的分页比较简单,只需要limit offset,length就可以获取数据了,但是当offset和length比较大的时候,mysql明显性能下降 1.子查询优化法 先找出第一条数据,然后大 ...
- CentOS 7 破解mariadb密码
1.停止mariadb服务: systemctl stop mariadb 2.进入单用户模式: mysqld_safe --skip-grant-tables & 3.切换到另外一个工作组 ...
- [Uva10641]Barisal Stadium(区间dp)
题意:按照顺时针给出操场的周边点,然后给出周围可以建设照明灯的位置,以及在该位置建设照明灯的代价,照明灯照射的范围与操场的边界相切,现在要求一个最小的花费,要求操场的所有边都被照射到. 解题关键:预处 ...