UVa 10051 Tower of Cubes(DP 最长序列)
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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
31 2 2 2 1 23 3 3 3 3 33 2 1 1 1 1101 5 10 3 6 52 6 7 3 6 95 7 3 2 1 91 3 3 5 8 106 6 2 2 4 41 2 3 4 5 610 9 8 7 6 56 1 2 3 4 71 2 3 3 2 13 2 1 1 2 30
Sample Output
Case #122 front3 front Case #281 bottom2 back3 right4 left6 top8 front9 front10 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另一个立方体且这个立方体底面的颜色等于另一个立方体顶面的颜色 这个立方体就可以放在另一个立方体上面 求这些立方体堆起来的最大高度;
每个立方体有6种放置方式 为了便于控制 可以将一个立方体分解为6个 每个立方体只用考虑顶面和底面 ;
d[i]表示分解后以第i个立方体为基底可以达到的最大高度 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;}
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