[POJ2392]Space Elevator

最新推荐文章于 2020-04-17 22:34:57 发布

dxyinme

最新推荐文章于 2020-04-17 22:34:57 发布

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分类专栏： --------------------动态规划背包型动态规划

本文链接：https://blog.youkuaiyun.com/dxyinme/article/details/81459977

--------------------动态规划同时被 2 个专栏收录

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背包型动态规划

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本文介绍了一个有趣的构建问题：如何利用不同类型的箱子搭建最高的太空电梯。文章详细解释了问题背景及限制条件，并提供了一种结合可行性背包动态规划与贪心策略的解决方案。

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Description

The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have $K (1 <= K <= 400)$ different types of blocks with which to build the tower. Each block of type i has height $h_i (1 <= h_i <= 100)$ and is available in quantity $c_i (1 <= c_i <= 10)$ . Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude $a_i (1 <= a_i <= 40000)$ .

Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.

Input

Line 1: A single integer, $K$
Lines $2..K+1$ : Each line contains three space-separated integers: $h_i, a_i$ , and $c_i$ . Line $i+1$ describes block type $i$ .

Output

Line 1: A single integer H, the maximum height of a tower that can be built

Sample Input

Sample Output

Hint
OUTPUT DETAILS:
From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.

题意:
给 $n$ 种箱子，第 $i$ 种箱子有 $c_i$ 个，箱子的高度是 $h_i$ ，第 $i$ 种箱子不能放在超过 $a_i$ 高度的地方。请问用这n种箱子最多可以搭多高的塔。

题解:
可行性背包DP+贪心。我们可以知道，如果我们确定了要用每种箱子用多少个，那么我们会尽量让 $a_i$ 比较小的箱子放在下面。那么我们就先把箱子按照 $a_i$ 排序。然后做一个可行性DP:

f[0]=1
for i  = 1 -> n
    for j=1 -> c[i]
        for k = a[i] downto h[i]  
            f[k] |= f[k-h[i]]

然后 $f[i]==1$ 就说明可以搭到的高度

#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define LiangJiaJun main
#define ll long long
using namespace std;
int n;
struct Ds{
    int a,h,c;
}d[404];
int f[40004];
inline bool dex(Ds A,Ds B){
    return A.a<B.a;
}
int w33ha(){
    memset(f,0,sizeof(f));
    int sum=0;
    for(int i=1;i<=n;i++){
        scanf("%d%d%d",&d[i].h,&d[i].a,&d[i].c);
        sum+=d[i].h*d[i].c;
    }
    sort(d+1,d+n+1,dex);
    f[0]=1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=d[i].c;j++){
            for(int k=min(sum,d[i].a);k>=d[i].h;k--){
                if(f[k-d[i].h]){
                    f[k]=1;
                }
            }
        }
    }
    for(int i=40000;i>=0;i--){
        if(f[i])return printf("%d\n",i),0;
    }
    return 0;
}
int LiangJiaJun (){
    while(scanf("%d",&n)!=EOF)w33ha();
    return 0;
}