标签：ems blocks sea test group determine height pes open
InputThe input file will contain one or more test cases. The first line of each test case contains an integer n,
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.
OutputFor each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".
Sample Input
1
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0
Sample Output
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
题意：有 n 种砖，每一种都无限多，分别有长宽高，可以任意摆放。现在要把这些砖叠起来，如果 A 放在 B 上，要求 A 的长宽比 B 的长宽都要小。问最多能叠多高。。。
思路：由于每种砖有无限多，直接枚举一块砖所有可能摆放方式的长宽高，然后按照长，宽排序 ... 然后就类似LIS的问题 ...
#include<bits/stdc++.h> #define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl #define pii pair<int,int> #define clr(a,b) memset((a),b,sizeof(a)) #define rep(i,a,b) for(int i = a;i < b;i ++) #define pb push_back #define MP make_pair #define LL long long #define ull unsigned LL #define ls i << 1 #define rs (i << 1) + 1 #define INT(t) int t; scanf("%d",&t) using namespace std; struct xx{ int c,k,g; xx(int c = 0,int k = 0,int g = 0): c(c),k(k),g(g){} }zhuan[185]; int dp[185]; bool cmp(xx A,xx B){ if(A.c != B.c) return A.c < B.c; return A.k < B.k; } int main() { int n; int cas = 0; while(~scanf("%d",&n) && n){ int cnt = 0; for(int i = 1;i <= n;++ i){ int l,r,k; scanf("%d%d%d",&l,&r,&k); zhuan[++ cnt] = xx(l,r,k); zhuan[++ cnt] = xx(l,k,r); zhuan[++ cnt] = xx(r,l,k); zhuan[++ cnt] = xx(r,k,l); zhuan[++ cnt] = xx(k,l,r); zhuan[++ cnt] = xx(k,r,l); } sort(zhuan + 1,zhuan + cnt + 1,cmp); rep(i,1,cnt + 1) dp[i] = zhuan[i].g; for(int i = 1;i <= cnt;++ i){ for(int j = i + 1;j <= cnt;++ j) if(zhuan[i].c < zhuan[j].c && zhuan[i].k < zhuan[j].k) dp[j] = max(dp[j],dp[i] + zhuan[j].g); } int maxx = 0; for(int i = 1;i <= cnt;++ i) maxx = max(maxx,dp[i]); cout << "Case " << ++ cas << ": maximum height = " << maxx << endl; // cout << maxx << endl; } return 0; }
Monkey and Banana HDU - 1069 （基础dp）
标签：ems blocks sea test group determine height pes open
原文地址：https://www.cnblogs.com/rookie-acmer/p/11270193.html
