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
Solution:
#include <stdio.h>
#define BOXN 30
#define BOXD 10
struct BOX {
int D[BOXD];
} B[BOXN];
int BI[BOXN], BD[BOXN], LEN[BOXN], K, N, L, k, n, l, d, V;
int main(){
while(scanf("%d %d", &K, &N) == 2){
/*Input Each Test Case*/
for(k = 0; k < K; k++){
BI[k] = k;
for(n = 0; n < N; n++){
scanf("%d", &d);
BD[n] = d;
}
/*Sort of each sequence*/
for(n = 0; n < N; n++){
d = n;
for(l = n+1; l < N; l++)
if(BD[d] > BD[l])
d = l;
B[k].D[n] = BD[d];
BD[d] = BD[n];
}/*End Sorting*/
}/*End Input*/
/*Sort boxes according to dimension's length*/
for(k = 0; k < K; k++)
BD[k] = B[k].D[0];
/*Sort of each Box*/
for(k = 0; k < K; k++){
d = k;
for(l = k+1; l < K; l++)
if(BD[d] > BD[l])
d = l;
BD[d] = BD[k];
n = BI[d];
BI[d] = BI[k];
BI[k] = n;
}/*End Sorting*/
/*Algorithm of LIS*/
LEN[BI[0]] = 1;
L = 1;
for(k = 1; k < K; k++){
l = 1;
for(d = 0; d < k; d++){
for(n = 0; n < N && B[BI[d]].D[n] < B[BI[k]].D[n]; n++);
if(n == N)
if(LEN[BI[d]] >= l){
BD[BI[k]] = BI[d];
l = LEN[BI[d]]+1;
}
}
LEN[BI[k]] = l;
if(L < l){
V = BI[k];
L = l;
}
}
BI[0] = V;
for(l = 1; l < L; l++){
BI[l] = BD[V];
V = BD[V];
}
printf("%d\n", L);
for(l = L-1; l >= 0; l--)
printf("%d ", BI[l]+1);
printf("\n");
}
return 0;
}
as shown in the diagram below:
![\begin{figure}
\centering
\setlength{\unitlength}{0.0125in} %
\begin{picture}
(2...
...raisebox{0pt}[0pt][0pt]{$\bullet
\bullet \bullet$ }}}
\end{picture}
\end{figure}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_soYEvq3lwx2jwHbtW04k9lF1CXvfLi_t6bC3as-zYg3Kp3hmtgA6YZHWfUnXq6KDOxHjJafss-uccrvgl_YybhaLwNN-xDuokmiDf9Tp60cYH3BKM=s0-d)
where n is the number of blocks) should appear
followed immediately by a colon.
If there is at least a block
on it, the colon must be followed by one space, followed by a list of
blocks that appear stacked in that position with each block number
separated from other block numbers by a space. Don't put any trailing
spaces on a line.