Uvaoj 437 The Tower of Babylon simple DP

Uvaoj 437 The tower of Babylon simple DP with a given number of types of bricks to build a tower, a brick (x, Y, z) three dimensions to represent, you can rotate them arbitrarily to make one face the bottom surface, the other surface becomes high, the process of building the tower, The top two dimensions of the face of the brick must be smaller than the two dimensions of the bottom face, and the number of bricks for each type is infinitely greater. Because it can be rotated arbitrarily, each brick can be rotated to get six kinds of the bottom of the brick (temporarily ignore the rotation to get the same), so, for each type of brick by rotation to get six different bricks, and then DP on the line, Dp[i] is labeled I as the bottom of the brick to the height of the tower, Note that the type of bricks must be sorted from small to large (x, Y, z) before DP. The code is as follows:

/************************************************************************* > File name:437.cpp > Author:gwq &G T mail:gwq5210@qq.com > Created time:2014 December 19 Friday 20:10 17 seconds **************************************************** /#include <cmath> #include <ctime> #include <cctype> #include <climits> # Include <cstdio> #include <cstdlib> #include <cstring> #include <map> #include <set> #incl Ude <queue> #include <stack> #include <string> #include <vector> #include <sstream> # Include <iostream> #include <algorithm> #define INF (INT_MAX/10) #define CLR (arr, Val) memset (arr, Val, siz
EOF (ARR)) #define PB push_back #define SZ (a) ((int) (a). Size ()) using namespace Std;
typedef set<int> SI;
typedef vector<int> VI;
typedef map<int, Int> Mii;

typedef long Long LL;

Const double ESP = 1e-5;

	#define N + struct Node {int x, y, Z; void InpuT (void) {scanf ("%d%d%d", &x, &y, &z);
		int a[3] = {x, y, z};
		Sort (A, a + 3);
		x = a[0];
		y = a[1];
	z = a[2];
	} void output (void) {printf ("%d%d%d\n", X, Y, z);

}}node[n];
int dp[n];

Node Tmp[n];
		BOOL CMP (node u, node v) {if (u.x = = v.x) {if (u.y = = v.y) {return u.z < v.z;
		} else {return u.y < v.y;
	}} else {return u.x < v.x;
	}} int main (int argc, char *argv[]) {int n;
	int c = 0;
		while (scanf ("%d", &n)! = EOF) {if (n = = 0) {break;
		} for (int i = 0; i < n; ++i) {tmp[i].input ();
		} int len = 0;
			for (int i = 0; i < n; ++i) {node[len++] = Tmp[i];
			node[len].x = TMP[I].Y;
			Node[len].y = tmp[i].z;
			Node[len].z = tmp[i].x;
			len++;
			node[len].x = tmp[i].x;
			Node[len].y = tmp[i].z;
			Node[len].z = TMP[I].Y;
			len++;
			node[len].x = tmp[i].z;
			Node[len].y = tmp[i].x;
			Node[len].z = TMP[I].Y;
			len++;
			node[len].x = tmp[i].z;
			Node[len].y = TMP[I].Y; NODE[LEN].Z = TMp[i].x;
			len++;
			node[len].x = TMP[I].Y;
			Node[len].y = tmp[i].x;
			Node[len].z = tmp[i].z;
		len++;
		}//for (int i = 0; i < len; ++i) {//node[i].output ();
		}//printf ("-------------\ n");
		Sort (node, node + len, CMP);
		for (int i = 0; i < len; ++i) {//node[i].output ();
		}//printf ("-------------\ n");
		CLR (DP, 0);
		Dp[0] = node[0].z;
			for (int i = 1; i < Len; ++i) {dp[i] = node[i].z; for (int j = 0; J < i; ++j) {if (node[j].x < node[i].x && Node[j].y < node[i].y && Dp[i] &lt ;
				DP[J] + node[i].z) {Dp[i] = Dp[j] + node[i].z;
		}}} int res = 0;
		for (int i = 0; i < len; ++i) {res = max (res, dp[i]);
	} printf ("Case%d:maximum height =%d\n", ++c, RES);
} return 0; }