Uva--10051（动规，记忆化搜索）

2014-08-02 19:37:06

  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 ( $1 \le N \le 500$) indicating the number of cubes you are given. The i­th ( $1 \le i \le N$) 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 

思路：又是一个经典的方块叠塔，但是这里要记录一下路径（用next[]数组），由于方块可以旋转（每个面有各自颜色），所以把一个方块变成6个（6种面朝上），但是要注意这和以前每个方块无限量有区别，每个方块只有一个，所以应该给每个方块定义一个权，使一个方块变出的6个方块有相同的权，以免对一个方块重复取用（这题恰好用重量来定义这个权）
  1 /*************************************************************************
  2     > File Name: k.cpp
  3     > Author: Nature
  4     > Mail: 564374850@qq.com
  5     > Created Time: Thu 31 Jul 2014 03:41:44 PM CST
  6 ************************************************************************/
  7 
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <vector>
 13 #include <iostream>
 14 #include <algorithm>
 15 using namespace std;
 16 
 17 struct node{
 18     int top;
 19     int bot;
 20     int w;
 21     int cnt;
 22     vector<int> s;
 23 }no[3005];
 24 
 25 int n,cnt;
 26 int dp[3005];
 27 int next[3005];
 28 
 29 void Build_graph(){
 30     for(int i = 0; i < cnt; ++i)
 31         for(int j = i + 1; j < cnt; ++j)
 32             if(no[i].w < no[j].w && no[i].bot == no[j].top){
 33                 ++no[i].cnt;
 34                 no[i].s.push_back(j);
 35             }
 36 }
 37 
 38 int Solve(int p){
 39     if(dp[p] != -1)
 40         return dp[p];
 41     int tmax = 0;
 42     int tem;
 43     for(int i = 0; i < no[p].cnt; ++i){
 44         if((tem = Solve(no[p].s[i])) > tmax){
 45             tmax = tem;
 46             next[p] = no[p].s[i];
 47         }
 48     }
 49     return dp[p] = tmax + 1;
 50 }
 51 
 52 int main(){
 53     //freopen("in","r",stdin);
 54     int Case = 0;
 55     int co[6];
 56     char str[6][10] = {"front","back","left","right","top","bottom"};
 57     while(scanf("%d",&n) == 1 && n){
 58         memset(dp,-1,sizeof(dp));
 59         memset(next,-1,sizeof(next));
 60         memset(no,0,sizeof(no));
 61         cnt = 6 * n;
 62         for(int i = 0; i < n; ++i){
 63             scanf("%d%d%d%d%d%d",&co[0],&co[1],&co[2],&co[3],&co[4],&co[5]);
 64             int k = i * 6;
 65             no[k].top = co[0];
 66             no[k].bot = co[1];
 67             no[k].w = i; // front
 68             no[k + 1].top = co[1];
 69             no[k + 1].bot = co[0];
 70             no[k + 1].w = i; //back
 71 
 72             no[k + 2].top = co[2];
 73             no[k + 2].bot = co[3];
 74             no[k + 2].w = i; //left
 75             no[k + 3].top = co[3];
 76             no[k + 3].bot = co[2];
 77             no[k + 3].w = i; //right
 78 
 79             no[k + 4].top = co[4];
 80             no[k + 4].bot = co[5];
 81             no[k + 4].w = i; //top
 82             no[k + 5].top = co[5];
 83             no[k + 5].bot = co[4];
 84             no[k + 5].w = i; //bottom
 85         }
 86         Build_graph();
 87         if(Case)
 88             printf("\n");
 89         printf("Case #%d\n",++Case);
 90         int ans = 0,tem,st;
 91         for(int i = 0; i < cnt; ++i){
 92             if((tem = Solve(i)) > ans){
 93                 ans = tem;
 94                 st = i;
 95             }
 96         }
 97         //for(int i = 0; i < cnt; ++i){
 98            // printf("dp[%d] : %d\n",i,dp[i]);
 99         //}
100         //for(int i = 0; i < cnt; ++i){
101            // printf("next[%d] : %d\n",i,next[i]);
102         //}
103         printf("%d\n",ans);
104         printf("%d %s\n",(st + 6) / 6,str[(st) % 6]);
105         while(next[st] != -1){
106             printf("%d %s\n",(next[st] + 6) / 6,str[(next[st]) % 6]);
107             st = next[st];
108         }
109     }
110     return 0;
111 }

 







            

            
posted @ 
2014-08-02 19:42 
Naturain 
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