HDU 1069 Monkey and Banana dp类型：最长上升子序列

http://acm.hdu.edu.cn/showproblem.php?pid=1069

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 14484    Accepted Submission(s): 7619

Problem Description

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

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：给你一些规格的箱子（无限个），现在让你把箱子堆起来，求最高能有多高。上面的箱子长宽都要严格小于下面的箱子。

思路：一个箱子有6种放法，排序后就模拟最长递增子序列的dp方法就行。dp入门题。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int INF = 1e8;
const int inf = 0x3f3f3f3f;
const int maxx = 10100;
const int N = 1000;
int n,cnt;
struct node
{
    int x,y,z;
} a[N];
int cmp(node s1,node s2)
{
    if(s1.x==s2.x)
        return s1.y>s2.y;
    return s1.x>s2.x;
}
int dp[N];
int ddp()
{
    sort(a,a+cnt,cmp);
    memset(dp,0,sizeof(dp));
    int num=0;
    for(int i=0; i<cnt; i++)
    {
        dp[i]=a[i].z;
        for(int j=0; j<i; j++)
        {
            if(a[i].x<a[j].x&&a[i].y<a[j].y&&a[i].z+dp[j]>dp[i])
                dp[i]=a[i].z+dp[j];
        }
        if(dp[i]>num)
            num=dp[i];
    }
    return num;
}
void add(int x,int y,int z)
{
    a[cnt].x=x,a[cnt].y=y,a[cnt++].z=z;
}
int main()
{
    int cas=1;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        int b,c,d;
        cnt=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d%d",&b,&c,&d);
            add(b,c,d);
            add(c,b,d);
            add(c,d,b);
            add(d,c,b);
            add(d,b,c);
            add(b,d,c);
        }
        printf("Case %d: maximum height = %d\n",cas++,ddp());
    }
    return 0;
}