poj[2392]space elevator
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.
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.
* 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
3 7 40 3 5 23 8 2 52 6
Sample Output
48
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.
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.
题解
题目大意:奶牛要去太空。为了实现这一目标,它们计划用k种石头建造一个高塔。每种石头有c个,每个高度为h,由于石头本身的质量问题,它们被放置的高度有一个限度,不能放在这一高度以上,记为a。求高塔可以达到的最大高度。
简单多重背包
#include<iostream> #include<algorithm> using namespace std; int n; bool f[40001]; struct block{ int h,c,a; bool operator<(const block ano)const{ return a<ano.a; } }b[401]; int main(){ ios::sync_with_stdio(false); cin>>n; for(int i=1;i<=n;i++) cin>>b[i].h>>b[i].a>>b[i].c; sort(b+1,b+1+n); f[0]=1; for(int i=1;i<=n;i++) for(int k=1;k<=b[i].c;k++) for(int j=b[i].a;j>=k*b[i].h;j--) if(f[j-b[i].h]) f[j]=1; for(int j=40000;j>=0;j--) if(f[j]){ cout<<j<<endl; break; } return 0; }