DP-HDU-1069-Monkey and Banana

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10168 Accepted Submission(s): 5285

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

就是个垒箱子的题，明显的DP。
每次输入一个箱子的数据时，分别以x，y，z为高存三个数据到箱子数据组里。
然后h[i]表示以第i个箱子数据为最高的箱子，所能达到的最高高度。
s是长宽能容纳第i个箱子数据的箱子。
那么动态方程就是h[i]=max(h[i],h[s]+h[i])。
数据比较水，n最大不超过30，所以n^2都能过。

////  main.cpp//  基础DP1-C-Monkeys and Banana////  Created by 袁子涵 on 15/10/23.//  Copyright © 2015年 袁子涵. All rights reserved.////  0ms 1540KB#include <iostream>#include <string.h>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <cmath>using namespace::std;typedef struct blocks{    int x,y,z;}Blocks;Blocks s[120];int n,h[120];bool visit[120];int dp(int num){    if (visit[num]) {        return h[num];    }    int x=s[num].x,y=s[num].y,z=s[num].z;    h[num]=z;    for (int i=1; i<=3*n; i++) {        if ((x<s[i].x && y<s[i].y) || (x<s[i].y && y<s[i].x)) {            h[num]=max(h[num],dp(i)+z);        }    }    visit[num]=1;    return h[num];}int main(int argc, const char * argv[]) {    int x,y,z,out,t=0;    while (scanf("%d",&n)!=EOF && n!=0) {        t++;        out=0;        for (int i=1; i<=3*n; i+=3) {            scanf("%d%d%d",&x,&y,&z);            s[i].x=x;s[i].y=y;s[i].z=z;            s[i+1].x=x;s[i+1].y=z;s[i+1].z=y;            s[i+2].x=y;s[i+2].y=z;s[i+2].z=x;        }        memset(h, 0, sizeof(h));        memset(visit, 0, sizeof(visit));        for (int i=1; i<=3*n; i++) {            out=max(out,dp(i));        }        printf("Case %d: maximum height = %d\n",t,out);    }    return 0;}