ZOJ 1093 Monkey and Banana

F - Monkey and Banana

Time Limit:2000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

SubmitStatusPracticeZOJ 1093

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 (x_i, y_i, z_i). 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 Specification

The input file will contain one or more test cases. The first line of each test case contains an integern,
representing the number of different blocks in the following data set. The maximum value forn is 30.
Each of the next n lines contains three integers representing the valuesx_i, y_i and z_i.
Input is terminated by a value of zero (0) for n.

Output Specification

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 "Casecase: maximum height = height"

Sample Input

110 20 3026 8 105 5 571 1 12 2 23 3 34 4 45 5 56 6 67 7 7531 41 5926 53 5897 93 2384 62 6433 83 270

Sample Output

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

DP+贪心。

学DP的时候的一道练习题。当时题解里说，N^3枚举每个面为底的情况，然后求最长上升子序列。

现在看来只要写结构体，然后排序就行。优先让长和宽小的放在前面。

之后就是DP，求最长上升子序列。

#include <stdio.h>#include <algorithm>#define N 35using namespace std;typedef struct{        int x,y,z;}Node;Node a[N*6];int n;int dp[N*6];bool cmp(Node a,Node b){        if(a.x==b.x)                if(a.y==b.y)                        return a.z>b.z;                else                        return a.y<b.y;        else                return a.x<b.x;}int main(){        int c=1;        while(scanf("%d",&n)>0&&n)        {                int k=0,maxn=0;                for(int i=0;i<n;i++)                {                        int x,y,z;                        scanf("%d%d%d",&x,&y,&z);                        a[k].x=x,a[k].y=y,a[k].z=z,k++;                        a[k].x=x,a[k].y=z,a[k].z=y,k++;                        a[k].x=y,a[k].y=x,a[k].z=z,k++;                        a[k].x=y,a[k].y=z,a[k].z=x,k++;                        a[k].x=z,a[k].y=y,a[k].z=x,k++;                        a[k].x=z,a[k].y=x,a[k].z=y,k++;                }                sort(a,a+k,cmp);                for(int i=0;i<k;i++)                        dp[i]=a[i].z;                for(int i=1;i<k;i++)                        for(int j=0;j<i;j++)                                if(a[i].x>a[j].x&&a[i].y>a[j].y)                                {                                        dp[i]=max(dp[j]+a[i].z,dp[i]);                                        maxn=max(maxn,dp[i]);                                }                printf("Case %d: maximum height = %d\n",c++,maxn);        }        return 0;}