标签：des style class blog c code
Some concepts in Mathematics and Computer Science are simple in one or two dimensions but become more complex when extended to arbitrary dimensions. Consider solving differential equations in several dimensions and analyzing the topology of an n-dimensional hypercube. The former is much more complicated than its one dimensional relative while the latter bears a remarkable resemblance to its ``lower-class‘‘ cousin.
Consider an n-dimensional ``box‘‘ given by its dimensions. In two dimensions the box (2,3) might represent a box with length 2 units and width 3 units. In three dimensions the box (4,8,9) can represent
a box 
(length, width, and height). In 6 dimensions it is, perhaps, unclear what the box (4,5,6,7,8,9) represents; but we can analyze
properties of the box such as the sum of its dimensions.
In this problem you will analyze a property of a group of n-dimensional boxes. You are to determine the longest nesting string of boxes, that is a sequence of boxes 
such
that each box 
nests in box
( 
.
A box D = ( 
) nests in a box E = ( 
)
if there is some rearrangement of the 
such that when rearranged each dimension is less than the corresponding dimension in box
E. This loosely corresponds to turning box D to see if it will fit in box E. However, since any rearrangement suffices, box D can be contorted, not just turned (see examples below).
For example, the box D = (2,6) nests in the box E = (7,3) since D can be rearranged as (6,2) so that each dimension is less than the corresponding dimension in E. The box D = (9,5,7,3) does NOT nest in the box E = (2,10,6,8) since no rearrangement of D results in a box that satisfies the nesting property, but F = (9,5,7,1) does nest in box E since F can be rearranged as (1,9,5,7) which nests in E.
Formally, we define nesting as follows: box D = ( ) nests in box E = ( )
if there is a permutation 
of 
such
that ( 
) ``fits‘‘ in ( )
i.e., if 
for all 
.
The input consists of a series of box sequences. Each box sequence begins with a line consisting of the the number of boxes k in the sequence followed by the dimensionality of the boxes, n (on the same line.)
This line is followed by k lines, one line per box with the n measurements of each box on one line separated by one or more spaces. The 
line
in the sequence ( 
) gives the measurements for the box.
There may be several box sequences in the input file. Your program should process all of them and determine, for each sequence, which of the k boxes determine the longest nesting string and the length of that nesting string (the number of boxes in the string).
In this problem the maximum dimensionality is 10 and the minimum dimensionality is 1. The maximum number of boxes in a sequence is 30.
For each box sequence in the input file, output the length of the longest nesting string on one line followed on the next line by a list of the boxes that comprise this string in order. The ``smallest‘‘ or ``innermost‘‘ box of the nesting string should be listed first, the next box (if there is one) should be listed second, etc.
The boxes should be numbered according to the order in which they appeared in the input file (first box is box 1, etc.).
If there is more than one longest nesting string then any one of them can be output.
5 2 3 7 8 10 5 2 9 11 21 18 8 6 5 2 20 1 30 10 23 15 7 9 11 3 40 50 34 24 14 4 9 10 11 12 13 14 31 4 18 8 27 17 44 32 13 19 41 19 1 2 3 4 5 6 80 37 47 18 21 9
5 3 1 2 4 5 4 7 2 5 6
这道题有两种解决办法。第一种可以看做是最长增子串问题的变形，第二种情况这是DAG上的最短路径。我采用的是第一种方法。
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<fstream>
using namespace std;
#define MAXSIZE 1000
int k,n; //k表示有多少组
//int a[MAXSIZE][MAXSIZE];
vector<vector<int> > vec;
vector<vector<int> > old_vec; //用来记录没排序的串
int d[MAXSIZE]; //用来描述状态的长度
int result[MAXSIZE][MAXSIZE]; //用来描述结果包含哪几项
int match[MAXSIZE]; //用来记录新串在旧串中的位子
bool compare(vector<int> vec1,vector<int> vec2)
{
int i;
for(i=0;i<n;i++)
{
if(vec1[i]==vec2[i])
continue;
return vec1[i]<vec2[i];
}
return false;
}
bool IFTRUE(vector<int> vec1,vector<int> vec2) //用来判断vec1嵌套在vec2中
{
int i;
for(i=0;i<n;i++)
if(vec1[i]>=vec2[i])
return false;
return true;
}
bool operator==(vector<int> vec1,vector<int> vec2)
{
unsigned int _size=vec1.size();
for(unsigned int i=0;i<_size;i++)
{
if(vec1[i]!=vec2[i])
return false;
}
return true;
}
int main()
{
//freopen("a.txt","r",stdin);
//fstream fcout("b.txt");
int i,j,temp,max_size,flag,m;
while(cin>>k)
{
cin>>n;
for(i=0;i<k;i++) //a[0]应当去掉，题目要求从1开始计算
{
vector<int> it;
for(j=0;j<n;j++)
{
cin>>temp;
it.push_back(temp);
}
sort(it.begin(),it.end()); //将box内元素升序排列
old_vec.push_back(it);
vec.push_back(it);
}
//将整个box串重新排序，这样有利于转化为最长增子串问题,同时还要记录新串与原串之间的对应关系
sort(vec.begin(),vec.end(),compare);
for(i=0;i<k;i++)
for(j=0;j<k;j++)
if(vec[i]==old_vec[j])
{
match[i]=j;
break;
}
//准备工作已做完，现在开始模拟最长增子序列
d[0]=1;
result[0][0]=0;
max_size=1;
flag=0;
for(i=1;i<k;i++)
{
d[i]=1;
for(j=0;j<i;j++)
{
if(IFTRUE(vec[j],vec[i])&&d[j]+1>d[i])
{
d[i]=d[j]+1;
for(m=0;m<d[j];m++)
result[i][m]=result[j][m];
result[i][m]=i;
}
}
if(d[i]>max_size)
{
max_size=d[i];
flag=i;
}
}
cout<<max_size<<endl;
for(i=0;i<max_size;i++)
{
if(i!=max_size-1)
cout<<match[result[flag][i]]+1<<" ";
else
cout<<match[result[flag][i]]+1<<endl;
}
vec.clear();
old_vec.clear(); //必须删除掉，不然会影响下一组数据
}
return 0;
}
| Stacking Boxes |
Some concepts in Mathematics and Computer Science are simple in one or two dimensions but become more complex when extended to arbitrary dimensions. Consider solving differential equations in several dimensions and analyzing the topology of an n-dimensional hypercube. The former is much more complicated than its one dimensional relative while the latter bears a remarkable resemblance to its ``lower-class‘‘ cousin.
Consider an n-dimensional ``box‘‘ given by its dimensions. In two dimensions the box (2,3) might represent a box with length 2 units and width 3 units. In three dimensions the box (4,8,9) can represent
a box (length, width, and height). In 6 dimensions it is, perhaps, unclear what the box (4,5,6,7,8,9) represents; but we can analyze
properties of the box such as the sum of its dimensions.
In this problem you will analyze a property of a group of n-dimensional boxes. You are to determine the longest nesting string of boxes, that is a sequence of boxes such
that each box nests in box ( .
A box D = ( ) nests in a box E = ( )
if there is some rearrangement of the such that when rearranged each dimension is less than the corresponding dimension in box
E. This loosely corresponds to turning box D to see if it will fit in box E. However, since any rearrangement suffices, box D can be contorted, not just turned (see examples below).
For example, the box D = (2,6) nests in the box E = (7,3) since D can be rearranged as (6,2) so that each dimension is less than the corresponding dimension in E. The box D = (9,5,7,3) does NOT nest in the box E = (2,10,6,8) since no rearrangement of D results in a box that satisfies the nesting property, but F = (9,5,7,1) does nest in box E since F can be rearranged as (1,9,5,7) which nests in E.
Formally, we define nesting as follows: box D = ( ) nests in box E = ( )
if there is a permutation of such
that ( ) ``fits‘‘ in ( )
i.e., if for all .
The input consists of a series of box sequences. Each box sequence begins with a line consisting of the the number of boxes k in the sequence followed by the dimensionality of the boxes, n (on the same line.)
This line is followed by k lines, one line per box with the n measurements of each box on one line separated by one or more spaces. The line
in the sequence ( ) gives the measurements for the box.
There may be several box sequences in the input file. Your program should process all of them and determine, for each sequence, which of the k boxes determine the longest nesting string and the length of that nesting string (the number of boxes in the string).
In this problem the maximum dimensionality is 10 and the minimum dimensionality is 1. The maximum number of boxes in a sequence is 30.
For each box sequence in the input file, output the length of the longest nesting string on one line followed on the next line by a list of the boxes that comprise this string in order. The ``smallest‘‘ or ``innermost‘‘ box of the nesting string should be listed first, the next box (if there is one) should be listed second, etc.
The boxes should be numbered according to the order in which they appeared in the input file (first box is box 1, etc.).
If there is more than one longest nesting string then any one of them can be output.
5 2 3 7 8 10 5 2 9 11 21 18 8 6 5 2 20 1 30 10 23 15 7 9 11 3 40 50 34 24 14 4 9 10 11 12 13 14 31 4 18 8 27 17 44 32 13 19 41 19 1 2 3 4 5 6 80 37 47 18 21 9
5 3 1 2 4 5 4 7 2 5 6
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标签：des style class blog c code
原文地址：http://blog.csdn.net/netuml/article/details/26666607
