UVa 103

　　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 = 0; i < n; i++)
        if (box1[i] >= box2[i])   return false;
    return true;
}

int dp(int i)    // compute d[i]
{
    if (d[i] > 0)   return d[i];
    d[i] = 1;
    for (int j = 1; j <= k; j++)
        if (G[i][j])
            d[i] = max(d[i], dp(j)+1);
    return d[i];
}

void print_path(int i)    // print the longest path starting with node i
{
    printf("%d ", i);
    for (int j = 1; j <= k; j++)
        if (G[i][j] && d[i] == d[j]+1)
        {
            print_path(j);
            break;
        }
}

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    while (scanf("%d%d", &k, &n) != EOF)
    {
        for (int i = 1; i <= k; i++)
            for (int j = 0; j < n; j++)
                scanf("%d", &box[i][j]);
        for (int i = 1; i <= k; i++)
            sort(box[i], box[i]+n);
        memset(G, 0, sizeof(G));
        for (int i = 1; i <= k; i++)
            for (int j = 1; j <= k; j++)
                if (is_nested(box[i], box[j]))
                    G[i][j] = 1;
        int ans = -1;
        int start = 0;
        memset(d, 0, sizeof(d));
        for (int i = 1; i <= k; i++)
        {
            int t = dp(i);
            if (t > ans)
            {
                ans = t;
                start = i;
            }
        }
        printf("%d
", ans);
        print_path(start);
        printf("
");
    }
    return 0;
}