UVa 103 - Stacking Boxes

　　题目大意：矩阵嵌套，不过维数是多维的。有两个个k维的盒子A(a₁, a₁...a_k), B(b₁, b₂...b_k)，若能找到(a₁...a_k)的一个排列使得a_i < b_i，则盒子A可嵌套在盒子B中。给出n个k维的盒子，找出最长的可嵌套的盒子的序列。实际上是DAG上的动态规划问题。首先是判断A能否嵌套在B中，对盒子的k维数进行排序，依次比较即可。然后用d[i]表示以节点i为起点的最长路径的长度，可以得到状态转移方程：d(i) = max{d(j)+1}, (i,j)是图上的一条边。最后就是打印路径，根据转移方程递归打印路径即可。

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 #define BOXN 35
 #define DIMENSION 12

 int k, n;    // k is the number of boxes, n is the dimensionality
 int box[BOXN][DIMENSION], d[BOXN];    // d[i] save the number of nodes in the longest path starting with node i
 int G[BOXN][BOXN];

 bool is_nested(int *box1, int *box2)
 {
     // if box1 can be nested in box2, return true; otherwise false
     for (int i = ; i < n; i++)
         if (box1[i] >= box2[i])   return false;
     return true;
 }

 int dp(int i)    // compute d[i]
 {
     if (d[i] > )   return d[i];
     d[i] = ;
     for (int j = ; j <= k; j++)
         if (G[i][j])
             d[i] = max(d[i], dp(j)+);
     return d[i];
 }

 void print_path(int i)    // print the longest path starting with node i
 {
     printf("%d ", i);
     for (int j = ; j <= k; j++)
         if (G[i][j] && d[i] == d[j]+)
         {
             print_path(j);
             break;
         }
 }

 int main()
 {
 #ifdef LOCAL
     freopen("in", "r", stdin);
 #endif
     while (scanf("%d%d", &k, &n) != EOF)
     {
         for (int i = ; i <= k; i++)
             for (int j = ; j < n; j++)
                 scanf("%d", &box[i][j]);
         for (int i = ; i <= k; i++)
             sort(box[i], box[i]+n);
         memset(G, , sizeof(G));
         for (int i = ; i <= k; i++)
             for (int j = ; j <= k; j++)
                 if (is_nested(box[i], box[j]))
                     G[i][j] = ;
         int ans = -;
         int start = ;
         memset(d, , sizeof(d));
         for (int i = ; i <= k; i++)
         {
             int t = dp(i);
             if (t > ans)
             {
                 ans = t;
                 start = i;
             }
         }
         printf("%d\n", ans);
         print_path(start);
         printf("\n");
     }
     return ;
 }