UVa 10051 Tower of Cubes(DP 最长序列)

Problem A: Tower of Cubes 

In this problem you are given N colorful cubes each having a distinct weight. Each face of a cube is colored with one color. Your job is to build a tower using the cubes you have subject to the following restrictions:

• Never put a heavier cube on a lighter one.
• The bottom face of every cube (except the bottom cube, which is lying on the floor) must have the same color as the top face of the cube below it.
• Construct the tallest tower possible.

Input 

The input may contain multiple test cases. The first line of each test case contains an integer N ( ) indicating the number of cubes you are given. The i­th ( ) of the next N lines contains the description of the i­th cube. A cube is described by giving the colors of its faces in the following order: front, back, left, right, top and bottom face. For your convenience colors are identified by integers in the range 1 to 100. You may assume that cubes are given in the increasing order of their weights, that is, cube 1 is the lightest and cube N is the heaviest.

The input terminates with a value 0 for N.

Output 

For each test case in the input first print the test case number on a separate line as shown in the sample output. On the next line print the number of cubes in the tallest tower you have built. From the next line describe the cubes in your tower from top to bottom with one description per line. Each description contains an integer (giving the serial number of this cube in the input) followed by a single white­space character and then the identification string (front, back, left, right, top or bottom) of the top face of the cube in the tower. Note that there may be multiple solutions and any one of them is acceptable.

Print a blank line between two successive test cases.

Sample Input 

3
1 2 2 2 1 2
3 3 3 3 3 3
3 2 1 1 1 1
10
1 5 10 3 6 5
2 6 7 3 6 9
5 7 3 2 1 9
1 3 3 5 8 10
6 6 2 2 4 4
1 2 3 4 5 6
10 9 8 7 6 5
6 1 2 3 4 7
1 2 3 3 2 1
3 2 1 1 2 3
0

Sample Output 

Case #1
2
2 front
3 front
 
Case #2
8
1 bottom
2 back
3 right
4 left
6 top
8 front
9 front
10 top

题意  给你n个立方体  立方体每面都涂有颜色  当一个立方体序号小于还有一个立方体且这个立方体底面的颜色等于还有一个立方体顶面的颜色  这个立方体就能够放在还有一个立方体上面  求这些立方体堆起来的最大高度；

每一个立方体有6种放置方式  为了便于控制  能够将一个立方体分解为6个  每一个立方体仅仅用考虑顶面和底面 ;

d[i]表示分解后以第i个立方体为基底能够达到的最大高度  j从1到i-1枚举  当满足top[i]==bot[j]时  d[i]=max(d[i],d[j]+1);

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 3005;
int buf[6], d[N], pre[N], n, m, ans;
char face[6][8] = {"front", "back", "left", "right", "top", "bottom"};
struct Cube{    int wei, bot, top;  } cube[N];

void print (int i)
{
    if (pre[i])  print (pre[i]);
    printf ("%d %s\n", cube[i].wei, face[ (i - 1) % 6]);
    return;
}

int main()
{
    int cas = 0;
    while (scanf ("%d", &n), n)
    {
        for (int i = m = 1; i <= n; ++i)
        {
            for (int k = 0; k < 6; ++k)
                scanf ("%d", &buf[k]);
            for (int k = 0; k < 6; ++k)
            {
                cube[m].wei = i;
                cube[m].top = buf[k];
                cube[m++].bot = (k % 2 ?
 buf[k - 1] : buf[k + 1]);
            }
        }
        memset (d, 0, sizeof (d));
        memset (pre, 0, sizeof (pre));

        for (int i = ans = 1; i <= 6 * n; ++i)
            for (int j = d[i] = 1; j < i; ++j)
                if (cube[j].wei < cube[i].wei && cube[j].bot == cube[i].top && d[i] < d[j] + 1)
                {
                    d[i] = d[j] + 1;
                    pre[i] = j;
                    if (d[i] > d[ans])
                        ans = i;
                }
        if (cas) printf ("\n");
        printf ("Case #%d\n%d\n", ++cas, d[ans]);
        print (ans);
    }
    return 0;
}