uva 103

 Stacking Boxes 

 

Background

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.

 

The Problem

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

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.

The Output

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.

 

Sample Input

 

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

 

Sample Output

 

5
3 1 2 4 5
4
7 2 5 6

解法：

本来只想到了暴力搜索，显然超时的嘛！然后想到用dp来做。

#include<iostream>

#include<cmath>

#define N 102

using namespace std;



int n,m,i,j,mx;



typedef struct

{

    int l[11];

    int index;

}ye;



ye pe[N];



int cmp(const void *a,const void *b)

{

    return *(int *)a-*(int *)b;

}



bool can(int x,int y)

{

    for(int k=0;k<m;k++)

        if(pe[x].l[k]>=pe[y].l[k]) return 0;

    return 1;

}



int leng[N],pro[N];





void print(int x)

{

    if(x==0)

        return;

    print(pro[x]);

    cout<<pe[x].index<<" ";

}







int main()

{

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        i=0;j=0;

        mx=0;

        for(i=1;i<=n;i++)

        {    

            for(j=0;j<m;j++)

            {    

                cin>>pe[i].l[j];

                pe[i].index=i;

            }

            qsort(pe[i].l,m,sizeof(int),cmp);//盒子内部排序

        }

        for(i=1;i<=n;i++)//所有盒子排序

        {

            for(j=i+1;j<=n;j++)

            {

                if(can(j,i))

                {

                    ye temp;

                    temp=pe[j];

                    pe[j]=pe[i];pe[i]=temp;

                }

            }

        }

        memset(leng,0,sizeof(leng));

        memset(pro,0,sizeof(pro));

        leng[1]=1;pro[1]=0;

        for(i=2;i<=n;i++)

        {

            leng[i]=1;

            for(j=i-1;j>=1;j--)

            {

                if(can(j,i))

                {

                    if(leng[i]<leng[j]+1)

                    {

                        leng[i]=leng[j]+1;

                        pro[i]=j;//记录下他放入的盒子

                    }

                }

            }

        }

        int mar=0;

        for(i=1;i<=n;i++)

            if(leng[i]>mx)

            {

                mx=leng[i];mar=i;

            }

        cout<<mx<<endl;

        print(pro[mar]);//递归输出盒子

        cout<<pe[mar].index;

        cout<<endl;

    }

    return 0;

}

递推公式也很简单。v[j]=max(v[i]+1|v[i]<v[j]),v[i]<v[j]表示i可以包含在j盒之内。
在dp之前有2个排序过程，一个是盒的内部的边从小到大排列，方便比较2个盒子是否可以放在一起。
一个是所有的盒子排列一遍，使所有前面的盒子都不可以把他后面的盒子放进去。
而对于输出的是哪几个盒子，又没有要求，所有中间只需记录下一级的盒子就可以了。
其他的地方就是输出的时候，不要多输出一个空格。