基础dp C Monkey and Banana （类最长上升子序列）

滴答滴答---题目链接

A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height.

They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.

Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.

Input

The 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.

Output

For 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

Sample Output

Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342

题目大意：有一个猴儿和一些种类的箱子，每种类箱子有各自的长宽高，数目有无限个。猴子想把箱子尽可能的堆积起来，

堆积的条件通俗的讲就是放在下面的箱子能够撑得住上面的箱子，按数学建模来说就是放在下面的箱子的长和宽要比放在上

面的要大（严格大于）；由于每种箱子虽然数量是无限的，但是肯定不能同种箱子直接累积，因为那样底面的边长就相等了，

这样就是不合法的了，所以每种箱子可以转换成六种情况的不同箱子（分别作为长宽高），之后就是个最长满足条件的高度序

列就行了，即套用最长上升子序列算法思想即可。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long int
#define INF 0x3f3f3f3f
#define Irish_Moonshine main
const int maxn = 205;
int dp[maxn];
struct node {int x, y, z;bool operator <(const node &t)const//对于底部长进行一波排序{return x > t.x;}
}blocks[maxn];
int cnt;
void add(int x, int y, int z)
{blocks[cnt].x = x;blocks[cnt].y = y;blocks[cnt].z = z;cnt++;
}
int Irish_Moonshine()
{int n, cs = 0;while (~scanf("%d", &n) && n){cnt = 0;int a, b, c;int ans = 0;for (int i = 1; i <= n; i++){scanf("%d%d%d", &a, &b, &c);add(a, b, c);add(a, c, b);add(b, a, c);add(b, c, a);add(c, a, b);add(c, b, a);}sort(blocks, blocks + cnt);memset(dp, 0, sizeof(dp));for (int i = 0; i < cnt; i++){dp[i] = blocks[i].z;//lowestfor (int j = 0; j < i; j++)//将高度最高的放置在i下面{if (blocks[j].x > blocks[i].x&&blocks[j].y > blocks[i].y)dp[i] = max(dp[i], dp[j] + blocks[i].z);//从j的位置上再放置一个i~~~}ans = max(dp[i], ans);}printf("Case %d: maximum height = %d\n", ++cs, ans);}return 0;
}

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