HDU 1069 Monkey and Banana

作者: ffacs 分类: 题目 发布时间: 2019-12-31 03:42

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

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 \(x_i\), \(y_i\) and \(z_i\). 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".

Solution

又是一个LCS,比板子复杂一点,要排个序。一个BOX放的形态可以有六种,每种z对应两种(x,y)。我们就把这六种\((x,y)\)和它对应的y都列出来。然后按\(x\)从小到大(大前提),\(y\)从小到大(小前提)来排序。这样lcs跑出来的肯定就是最优解了。然后跑一遍lcs的板子,就完事了

#include<bits/stdc++.h>
#define rg register
#define il inline
using namespace std;
typedef long long ll;
ll read(){
    ll ans = 0,flag = 1; char ch;while((ch=getchar())<'0'||ch>'9') if(ch =='-') flag=-1;ans = ch ^ 48;while((ch=getchar())>='0'&&ch<='9') ans=(ans<<3)+(ans<<1)+(ch^48);return flag*ans;}
void WRI(ll x){if(x<0){ putchar('-');x=-x; }if(x>9) WRI(x/10);putchar(x%10+'0');}
void write(int x,char o){ WRI(x),putchar(o); }
struct  box{
    int d[3];
}b[100000];
bool cmp(box a,box b){
    return a.d[0]==b.d[0]?a.d[1]<b.d[1]:a.d[0]<b.d[0];
}
int dp[100000];
int main(){
    int n,c=1;
    while(cin>>n&&n){
        int x,y,z,t=1;
        b[0].d[0]=b[0].d[1]=199999999;
        for(rg int i=1;i<=n;i++){
            cin>>x>>y>>z;
            b[t].d[0]=x,b[t].d[1]=y,b[t].d[2]=z;t++;
            b[t].d[0]=x,b[t].d[1]=z,b[t].d[2]=y;t++;
            b[t].d[0]=y,b[t].d[1]=x,b[t].d[2]=z;t++;
            b[t].d[0]=y,b[t].d[1]=z,b[t].d[2]=x;t++;
            b[t].d[0]=z,b[t].d[1]=x,b[t].d[2]=y;t++;
            b[t].d[0]=z,b[t].d[1]=y,b[t].d[2]=x;t++;
        }
        sort(b+1,b+t,cmp);
        int ans=0;
        for(rg int i=1;i<t;i++){
            dp[i]=b[i].d[2];
            for(rg int j=i;j>=1;j--){
                if(b[i].d[0]>b[j].d[0]&&b[i].d[1]>b[j].d[1])
                    ans=max(ans,dp[i]=max(dp[j]+b[i].d[2],dp[i]));
            }
        }
        printf("Case %d: maximum height = %d\n",c++,ans);
        memset(dp,0,sizeof dp);
    }
    return 0;
}

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