因为每种block都有无限多个，所以(x,y,z)等价于是(x,y,z)+(x,z,y)+(y,z,x)这三种block。
然后就像是矩形嵌套那样求解，只不过状态转移方程变成了：
dp[i] = max( dp[i], dp[j] + z[i] );
代码如下：
1 #include <algorithm> 2 #include <cstdio> 3 using namespace std; 4 5 const int N = 100; 6 int dp[N]; 7 8 struct Node 9 { 10 int x, y, z; 11 bool operator < ( const Node & o ) const 12 { 13 if ( x != o.x ) return x < o.x; 14 return y < o.y; 15 } 16 Node(){} 17 Node( int _x, int _y, int _z ) 18 { 19 x = _x, y = _y, z = _z; 20 } 21 void standard() 22 { 23 if ( x > y ) swap( x, y ); 24 } 25 } node[N]; 26 27 int main () 28 { 29 int n, _case = 1; 30 while ( scanf("%d", &n), n ) 31 { 32 for ( int i = 0; i < n; i++ ) 33 { 34 int x, y, z; 35 scanf("%d%d%d", &x, &y, &z); 36 node[i * 3] = Node( x, y, z ); 37 node[i * 3].standard(); 38 node[i * 3 + 1] = Node( x, z, y ); 39 node[i * 3 + 1].standard(); 40 node[i * 3 + 2] = Node( y, z, x ); 41 node[i * 3 + 2].standard(); 42 } 43 sort( node, node + 3 * n ); 44 int ans = -1; 45 for ( int i = 0; i < 3 * n; i++ ) 46 { 47 dp[i] = node[i].z; 48 for ( int j = 0; j < i; j++ ) 49 { 50 if ( node[j].x < node[i].x && node[j].y < node[i].y ) 51 { 52 dp[i] = max( dp[i], dp[j] + node[i].z ); 53 } 54 } 55 ans = max( ans, dp[i] ); 56 } 57 printf("Case %d: maximum height = %d ", _case++, ans); 58 } 59 return 0; 60 }