hdoj 1069 Monkey and Banana（上升子序列最大和）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12836    Accepted Submission(s): 6737

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
  

 

Source

University of Ulm Local Contest 1996

 

Recommend

JGShining

/*
每个木箱可以变成3个木箱
按长度递减排序，转换成最大上升子序列和
*/
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 50;
int dp[maxn*3];
struct node
{
    int x, y, h;
    node(int xx, int yy, int hh): x(xx), y(yy), h(hh) {}
    node() {}
    bool operator <(const node &a) const
    {
        if(x == a.x) return y < a.y;    //长一样的话宽递增排序，防止长度一样还能叠上去
        return x > a.x;
    }
}block[maxn*3];
int main(void)
{
    int n, ca = 1;
    while(cin >> n, n)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++)
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            block[i*3] = node(max(a, b), min(a, b), c);
            block[i*3+1] = node(max(a, c), min(a, c), b);
            block[i*3+2] = node(max(b, c), min(b, c), a);
        }
        sort(block, block+3*n);
        for(int i = 0; i < 3*n; i++) dp[i] = block[i].h;
//        for(int i = 0; i < 3*n; i++)
//            printf("%d %d %d\n", block[i].x, block[i].y, block[i].h);
        int ans = block[0].h;
        for(int i = 1; i < 3*n; i++)
            for(int j = 0; j < i; j++)
                if(block[j].y > block[i].y && dp[j]+block[i].h > dp[i])
                    dp[i] = dp[j]+block[i].h, ans = max(ans, dp[i]);
        printf("Case %d: maximum height = %d\n", ca++, ans);
    }
    return 0;
}