UVA 437 the Tower of Babylon

Test instructions: There are n cubes, to be piled into a master as high as possible, each cube is placed above the length of the cube is strictly less than below, each cube has an unlimited

Analysis: My approach is that each cube has a maximum of 6 states, then also have a 6n fixed length and width of the cube, then this is a DAG on the longest road, 6n Cube, O (n^2) algorithm,

Very simple, see the code for details

#include <bits/stdc++.h>using namespacestd;Const intmaxn=205;intM,N,DP[MAXN];structnode{intx, Y, Z voidInitintAintBintc) {x=a;y=b;z=C; }}P[MAXN];BOOLCMP (node A,node b) {returna.x*a.y<b.x*b.y;}BOOLRead () {scanf ("%d",&N); if(!n)return false; M=0;  while(n--){        intA,b,c; scanf ("%d%d%d",&a,&b,&c); P[m++].init (A,B,C); P[m++].init (B,A,C); P[m++].init (A,C,B); P[m++].init (b,c,a); P[m++].init (C,A,B); P[m++].init (c,b,a); } sort (P,p+m,cmp); return true;}voidsolve () {intans=0;  for(intI=0; i<m;i++) {Dp[i]=p[i].z;  for(intj=0; j<i;j++)          if(p[i].x>p[j].x&&p[i].y>p[j].y) Dp[i]=max (dp[i],dp[j]+p[i].z); Ans=Max (ans,dp[i]); } printf ("%d\n", ans);}intMain () {intcas=1;  while(Read ()) {printf ("Case %d:maximum height =", cas++);    Solve (); }    return 0;}

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Here is the code for Rujia

//UVa437 the Tower of Babylon//Rujia Liu//algorithm: The longest path on a dag with a status of (IDX, k), that is, the current top is a cube idx, where the K bar (after sorting) is high#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;#defineREP (i,n) for (int i = 0; I < (n); i++)Const intMAXN = -+5;intN, blocks[maxn][3], d[maxn][3];voidGet_dimensions (int* V,intBintDim) {intIDX =0; REP (i,3)if(I! = Dim) v[idx++] =blocks[b][i];}intdpintIintj) {int& ans =D[i][j]; if(Ans >0)returnans; Ans=0; intv[2], v2[2];  Get_dimensions (V, I, j); Rep (A,n) rep (b,3) {get_dimensions (v2, A, b); if(v2[0] < v[0] && v2[1] < v[1]) ans =Max (ans, DP (A, b)); } ans+=Blocks[i][j]; returnans;}intMain () {intKase =0;  while(SCANF ("%d", &n) = =1&&N) {rep (i,n) {Rep (J,3) scanf ("%d", &Blocks[i][j]); Sort (blocks[i], blocks[i]+3); } memset (d,0,sizeof(d)); intAns =0; Rep (I,n) Rep (J,3) ans =Max (ans, DP (I,J)); printf ("Case %d:maximum height =%d\n", ++Kase, ans); }  return 0;}

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UVA 437 the Tower of Babylon