UVa 10051-tower of Cubes

Title: Give you n cubes, each face has a color, they are arranged in a way of increasing weight, now require them to be lined up in a tower,

Each layer of a cube, requires each cube above the weight of the cube is less than he, but also to ensure that the contact surface of the same color,

Ask how many layers the highest can be placed, the answer is not unique.

Analysis: DP, dynamic programming, Lis. The maximum non-descending subsequence, which has been sequenced, satisfies the same surface color as the contact.

Defined state: F (i,k) is the maximum height of the K face up when the first block is the top;

Transfer equation: F (i,k) = Max (f (j,f)) {J < I, and the same face at the same place};

Each block is placed in 6 French, using a one-dimensional array compression state record precursor, output path.

Explanation: Go to opposite side direct different or operation can, long time did not brush the question ╮(╯▽╰)╭.

#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include < Cstdio> #include <cmath>using namespace std;int cubes[505][6];int length[505][6];int Front[3333];char face[6] [7] = {' Front ', ' Back ', ' left ', ' right ', ' top ', ' bottom '};void output (int space, int n) {if (space) {printf ("%d%s\n", N-spa CE/6,FACE[SPACE%6]); output (Front[space], n);}} int main () {int n,t = 0;while (~scanf ("%d", &n) && N) {for (int i = n; I >= 1;--i) {for (int j = 0; J < 6; + + j) {scanf ("%d", &cubes[i][j]);}} int max = 0,space = 0;for (int i = 1; I <= n; + + i) {for (int k = 0; k < 6; + + K) {length[i][k] = 1;front[6*i+k] = 0 ; for (int j = 1; j < I; + + j) {for (int f = 0; F < 6; + f) {if (cubes[j][f] = = cubes[i][k^1] && length[j][f ] >= Length[i][k]) {length[i][k] = length[j][f]+1;front[6*i+k] = 6*j+f;}}} if (Max < length[i][k]) {max = Length[i][k];space = 6*i+k;}}} if (t + +) printf ("\ n");p rintf ("Case #%d\n", T);p rintf ("%d\n ", max); output (space, n+1);} return 0;}

UVa 10051-tower of Cubes