UV 103 Stacking Boxes (DP), stackingboxes

UV 103 Stacking Boxes (DP), stackingboxes
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 topologyN-Dimen1_hypercube. The former is much more complicated than its one dimen1_relative while the latter bears a remarkable resesponance to its ''lower-class'' cousin.

The Problem

ConsiderN-Dimenstm''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 (,) 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 groupN-Dimen1_boxes. You are to determine the longestNesting stringOf 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 = () nests in the box E = () since D can be rearranged) so that each dimension is less than the corresponding dimension in E. the box D = (,) does NOT nest in the box E = (,) since no rearrangement of D results in a box that satisfies the nesting property, but F = (,) does nest in box E since F can be rearranged as (,) which nests in E.

Formally, we define nesting as follows: box D = ()NestsIn 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 number of boxesKIn the sequence followed by the dimensionality of the boxes,N(On the same line .)

This line is followedKLines, one line per box withNMeasurements 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 shocould process all of them and determine, for each sequence, which ofKBoxes 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 ''nermost'' box of the nesting string shocould be listed first, the next box (if there is one) shocould be listed second, etc.

The boxes shoshould 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 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 54
7 2 5 6
Dynamic Planning On DAG:
Question: the extension of rectangular nesting changes two sides into n faces.
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>typedef long long LL;using namespace std;const int maxn=110;int dp[maxn],mp[maxn][maxn],a[maxn][maxn];int n,m,ans,pos;int ok(int i,int j){    for(int k=0;k<m;k++)        if(a[i][k]>=a[j][k])   return 0;    return 1;}int dfs(int i){    if(dp[i])  return dp[i];    dp[i]=1;    for(int k=0;k<n;k++)        if(i!=k&&mp[k][i])  dp[i]=max(dp[i],dfs(k)+1);    return dp[i];}void print(int x){    for(int i=0;i<n;i++)    {        if(dp[i]==dp[x]-1&&mp[i][x])        {            print(i);            break;        }    }    if(x==pos)  printf("%d",x+1);    else  printf("%d ",x+1);}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(dp,0,sizeof(dp));        memset(mp,0,sizeof(mp));        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)                scanf("%d",&a[i][j]);            sort(a[i],a[i]+m);        }        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)               if(i!=j)   mp[i][j]=ok(i,j);        }        ans=0;        for(int i=0;i<n;i++)        {            int temp=dfs(i);            if(ans<temp)            {                ans=temp;                pos=i;            }        }        printf("%d\n",ans);        print(pos);        puts("");    }    return 0;}