UVA 103 (Stacking Boxes)

Topic

Description

Background

Some Concepts in Mathematics and computer science is simple in one or both dimensions but become more complex when Extende D to arbitrary dimensions. Consider solving differential equations in several dimensions and analyzing the topology of ann-dimensional Hyper Cube. The former is much more complicated than it one dimensional relative while the latter bears a remarkable resemblance to I Ts ' Lower-class ' cousin.

The problem

Consider an n-dimensional "box" given by its dimensions. In both 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 was, 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'll analyze a property of a group of n-dimensional boxes. You were to determine the longestnesting string of boxes, that's 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 then rearranged each dimension are Les s than the corresponding dimension in box E. This loosely corresponds-turning box D to see if it would fit in box E. However, since any rearrangement suffices, box D can is contorted, not just turned (see examples below).

For example, the box D = (2,6) Nests in the box E = (7,3) since D can being rearranged as (6,2) So, all dimension is Les s 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 is rearranged as (1,9,5,7) which nests in E.

Formally, we define nesting as follows:box D = ()nests in box E = () If there are a permutation of such that ( ) ' Fits ' in () i.e.

The Input

The input consists of a series of box sequences. Each box sequence begins with a line consisting of the the number of boxesK in the sequence followed by the Dimen Sionality of the boxes, n (on the same line.)

This are followed by K -lines, one line per box and the n measurements of each box on one line Separ Ated by one or more spaces. The sequence () gives the measurements for the box.

There may is several box sequences in the input file. Your program should process all of them and determine, for each sequence, which of theK boxes determine the longe St Nesting string and the length of the nesting string (the number of boxes in the string).

In this problem, the maximum dimensionality is, 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, and 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 a string then any one of the them can be output.

Sample Input

5 23 78 105 29 1121 188 65 2 20 1 30 1023 15 7 9 11 340 50 34 24 14 49 10 11 12 13 1431 4 18 8 27 1744 32 13 19 41 191 2 3 4 5 680 37 47 18 21 9

Sample Output

53 1 2 4 547 2 5 6

Test instructions

For n-dimensional graphics, their side length is {D1,D2,D3...DN}, and for two n-dimensional graphs, if all of the edge lengths of one of them correspond to an edge length that is less than the other in any order, then the lock can be nested to another. For example a{1,2}, b{2,3}, a all edge lengths correspond to the length of the edge less than B, so a can be nested in B.

For k-dimensional graphs, find out how many of them can be nested in a row.

"Thinking of the problem" memory search or the longest ascending subsequence sequence!!!

First of all to determine whether the two graphics can be nested, just want to all the graphics edge length in accordance with the order from the beginning, then for a, b two, as long as the order to compare their number of sides, if all ai<bi, so a can be nested in B.

Then, is the solution, the problem has two methods.

The first is the so-called DAG model with memory search (see "Algorithmic Primer Classic" p161).

We can use the adjacency matrix of graphs to represent all relationships, a "can be nested in" B, then it is g[a][b]=1. The problem can then be translated into what is the longest path length for this graph.

The second method, suppose a can be nested in B with A<b to express, then the longest string is a<b<c<d<e ..., you can see very much like a "maximum increment subsequence", where the "increment" refers to "dimension" increment.

"AC Code, Memory search version, longest ascending sub-sequence and this difference is not too much"

/**** Memory Search ****/#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm  >using namespace Std;int g[32][32],arr[32][12];int dis[32],n,k;bool Fuck;//judge arr[a] and arr[b] small or big.inline    BOOL Judge (int A,int b) {for (int i=0; i<k; i++) {if (Arr[a][i]<=arr[b][i]) return false; } return true;    int dfs (int i) {if (dis[i]!=-1) return dis[i];    int &ans=dis[i]=1;        for (int j=0; j<n; J + +) {if (G[i][j]) {ans= max (Ans,dfs (j) +1); }} return ans;    void Print_path (int pos) {if (fuck) printf ("%d", pos+1);        else {printf ("%d", pos+1);    Fuck=true;            } for (int j=0; j<n; J + +) {if (G[pos][j]&&dis[j]+1==dis[pos]) {Print_path (j);        Break }}}int Main () {while (~scanf ("%d%d", &n,&k)) {for (Int. i=0; i<n; i++) {for (in T j=0; j<k; J + +) {scanf("%d", &arr[i][j]);        } sort (arr[i],arr[i]+k);        } memset (G,0,sizeof (G));        for (int i=0, i<n-1; i++) {for (int j=i+1; j<n; J + +) {if (judge (I,j))                G[j][i]=1;            else if (judge (j,i)) g[i][j]=1;        }} memset (dis,-1,sizeof (dis));        int ans=-1,pos;            for (int i=0; i<n; i++) {int T=dfs (i);                if (T>ans) {ans=t;            Pos=i;        }} printf ("%d\n", ans);        Fuck=false;        Print_path (POS);    printf ("\ n"); } return 0;}

UVA 103 (Stacking Boxes)