UVA 103 Stacking Boxes (DAG)

UVA 103 Stacking Boxes

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 a n-dimensional hypercube. 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 tex2html_wrap_inline40 (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. Determine the longest nesting string of boxes, that's a sequence of boxes tex2html_wrap_inline44 such that EAC H box tex2html_wrap_inline46 nests in box tex2html_wrap_inline48 (Tex2html_wrap_inline50.

A box D = (tex2html_wrap_inline52) Nests in a box E = (tex2html_wrap_inline54) If there is some rearrangement of the T Ex2html_wrap_inline56 such that's when rearranged each dimension was less 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 = (tex2html_wrap_inline52) nests in box E = (tex2html_wrap_inline54) if There is a permutation tex2html_wrap_inline62 of tex2html_wrap_inline64 such this (TEX2HTML_WRAP_INLINE66) "fits" in (t EX2HTML_WRAP_INLINE54) i.e., if tex2html_wrap_inline70 for all tex2html_wrap_inline72.

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 Dimensionalit Y 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 separated by one or mor E spaces. The tex2html_wrap_inline82 line in the sequence (tex2html_wrap_inline84) gives the measurements for the tex2html_wrap_in Line82 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 the k boxes determine the longest nesti ng string and the length of this 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. smallest‘‘ orthe innermost "box of the nesting string should be listed first, the next box (if there are one) should be listed Seco nd, 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 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

The main topic: There are N m-side graphics, when each edge of a graph is greater than the other graph, you can nest this shape. Ask, the most nested way. The maximum number of nested arrays is output and the nesting is output from small to large. Problem-solving ideas: First the edge of each M-edge is sorted by size, then the DAG issue.

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#define N -using namespace STD;intN, M;intNum[n][n], Dp[n], rec[n];intCheckintAintb) {//Determine if polygon A can nest Polygon B     for(inti =0; I < m; i++) {if(Num[a][i] <= num[b][i]) {return 0; }    }return 1;}intDP (intx) {//Returns the nesting method that owns the polygon, the maximum number of nested layers    if(Dp[x])returnDP[X];//DP Array records, the maximum number of nested layers.      for(inti =0; I < n; i++) {if(Check (i, x)) {inttemp = DP (i);if(Temp > Dp[x])                {Dp[x] = temp;            REC[X] = i; }        }    }return++DP[X];}intSolve () {intMax =0, R; for(inti =0; I < n; i++) {if(DP (i) > Max)            {r = i;        Max = DP (i); }    }printf("%d\n", Max); for(inti =1; i < Max; i++) {printf("%d", R +1);     r = Rec[r]; }printf("%d\n", R +1);}intMain () { while(scanf("%d%d", &n, &m) = =2) { for(inti =0; I < n; i++) { for(intj =0; J < M; J + +) {scanf("%d", &num[i][j]);        } sort (Num[i], num[i] + m); }memset(DP,0,sizeof(DP));memset(REC,0,sizeof(rec));    Solve (); }return 0;}

UVA 103 Stacking Boxes (DAG)