hdoj1069 Monkey and Banana
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5829 Accepted Submission(s): 2961
group of researchers are designing an experiment to test the IQ of a
monkey. They will hang a banana at the roof of a building, and at the
mean time, provide the monkey with some blocks. If the monkey is clever
enough, it shall be able to reach the banana by placing one block on the
top another to build a tower and climb up to get its favorite food.
The
researchers have n types of blocks, and an unlimited supply of blocks
of each type. Each type-i block was a rectangular solid with linear
dimensions (xi, yi, zi). A block could be reoriented so that any two of
its three dimensions determined the dimensions of the base and the other
dimension was the height.
They want to make sure that the
tallest tower possible by stacking blocks can reach the roof. The
problem is that, in building a tower, one block could only be placed on
top of another block as long as the two base dimensions of the upper
block were both strictly smaller than the corresponding base dimensions
of the lower block because there has to be some space for the monkey to
step on. This meant, for example, that blocks oriented to have
equal-sized bases couldn't be stacked.
Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.
each test case, print one line containing the case number (they are
numbered sequentially starting from 1) and the height of the tallest
possible tower in the format "Case case: maximum height = height".
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct T
{
int x,y,z;
}c[];
int cmp(T a,T b)
{
if(a.x<b.x)
return ;
if(a.x==b.x && a.y<b.y)
return ;
return ;
}
int main()
{
int i,j,n,t,max,a[],k=,opt[];
while(cin>>t && t)
{j=;n=*t;max=;
memset(opt,,sizeof(opt));
while(t--)
{
for(i=;i<;i++)
cin>>a[i];
sort(a,a+);//如果不进行排序,下面就会有六种可能了,
c[j].x=a[];c[j].y=a[];c[j].z=a[];j++;//a[2]>a[1]>a[0]
c[j].x=a[];c[j].y=a[];c[j].z=a[];j++;//所以长始终大于或等于宽
c[j].x=a[];c[j].y=a[];c[j].z=a[];j++;//这样就减少了三种情况
}
sort(c,c+n,cmp);//先按长进行排序,然后对宽进行排序
for(i=;i<n;i++)
{opt[i]=c[i].z;
for(j=;j<i;j++)
{
if(c[i].x>c[j].x && c[i].y>c[j].y && opt[j]+c[i].z>opt[i])//状态转移,选了就加上;
opt[i]=opt[j]+c[i].z;
}
// cout<<"opt="<<opt[i]<<endl;
}
for(i=;i<n;i++)
if(opt[i]>max)
max=opt[i];
k++;
cout<<"Case "<<k<<": maximum height = "<<max<<endl;
}
return ;
}
hdoj1069 Monkey and Banana的更多相关文章
- hdoj1069 Monkey and Banana(DP--LIS)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1069 思路: 由题意,显然一种block可能有6种形式,且一种形式最多使用一次,因此最多有30×6=1 ...
- hdu 1069 Monkey and Banana
Monkey and Banana Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- 杭电oj 1069 Monkey and Banana 最长递增子序列
Monkey and Banana Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU 1069 Monkey and Banana(二维偏序LIS的应用)
---恢复内容开始--- Monkey and Banana Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K ...
- ACM-经典DP之Monkey and Banana——hdu1069
***************************************转载请注明出处:http://blog.csdn.net/lttree************************** ...
- HDU 1069 Monkey and Banana (DP)
Monkey and Banana Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u S ...
- HDU 1069 Monkey and Banana(动态规划)
Monkey and Banana Problem Description A group of researchers are designing an experiment to test the ...
- Monkey and Banana(HDU 1069 动态规划)
Monkey and Banana Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- ZOJ 1093 Monkey and Banana (LIS)解题报告
ZOJ 1093 Monkey and Banana (LIS)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...
随机推荐
- Effective C++ 38-42
38.绝不要又一次定义继承而来的缺省參数值. 又一次定义函数缺省參数值意味着又一次定义函数.而非虚函数不能又一次定义,所以将就考虑不能又一次定义虚函数的缺省參数值的原因:虚函数是动态绑定的而缺省參数值 ...
- Eclipse Console 加大显示的行数和禁止错误弹出
在 Preferences-〉Run/Debug-〉Console里边,去掉对Limit console output的选择,或者选择,设置一下buffer size的设定值 禁止弹出: Prefer ...
- plsql oracle client没有正确安装(plsql连接远程数据库)
plsql oracle client没有正确安装(plsql连接远程数据库) CreateTime--2018年4月23日16:55:11 Author:Marydon 1.情景再现 2.问题解 ...
- .NET 垃圾回收机制要点整理
1. .NET资源分托管资源和非托管资源,对于托管资源,.NET GC可以很好的回收无用的垃圾,而对于非托管(例如文件访问,网络访问等)需要手动清理垃圾(显式释放). 2. 非托管资源的释放,.NET ...
- 10、java初始化顺序
在new B一个实例时首先要进行类的装载.(类只有在使用New调用创建的时候才会被java类装载器装入) 2,在装载类时,先装载父类A,再装载子类B3,装载父类A后,完成静态动作(包括静态代码和变 ...
- java装箱拆箱
基本数据类型的自动装箱(autoboxing).拆箱(unboxing)是自J2SE 5.0开始提供的功能. 一般我们要创建一个类的对象的时候,我们会这样: Class a = new Class(p ...
- Linux下TCP最大连接数受限问题
一. 文件数限制修改1.用户级别查看Linux系统用户最大打开文件限制:# ulimit -n1024 (1) vi /etc/security/limits.confmysql soft nofil ...
- 升级 asp.net core 1.1 到 2.0 preview
Upgrading to .NET Core 2.0 Preview 1 更新 依赖的类库 改为 标准库 2 web app 更改 csproj 文件---升级版本 <PropertyGrou ...
- 开源大数据技术专场(下午):Databircks、Intel、阿里、梨视频的技术实践
摘要: 本论坛第一次聚集阿里Hadoop.Spark.Hbase.Jtorm各领域的技术专家,讲述Hadoop生态的过去现在未来及阿里在Hadoop大生态领域的实践与探索. 开源大数据技术专场下午场在 ...
- div 模糊效果
-webkit-filter:blur(3px); -moz-filter:blur(3px); filter:url(blur.svg#blur); filter: progid:DXImageTr ...