POJ 2392 Space Elevator（01背包）

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12063		Accepted: 5731

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

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.
Source

USACO 2005 March Gold

题目大意：
一群牛要上太空，给出n种石块，每种石块给出单块高度，总高度不能超过的最大值，数量，要求用这些石块能组成
的最大高度

解题思路：
先按高度最高值按从小到大排序，那么再运用01背包时可以得到最优结果
用dp[ i ]=1表示 i 高度是可以用石块组成的
sum[ i ]表示在用一种石块时，达到 i 高度需要的石块数
AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[40005],sum[40005];
struct node{
int h,a,c;
}f[405];
bool cmp(node a,node b){
if(a.a<b.a)return true;
return false;
}
int main(){
int k;
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d%d%d",&f[i].h,&f[i].a,&f[i].c);
}
sort(f,f+k,cmp);
int ans=0;
dp[0]=1;
for(int i=0;i<k;i++){
memset(sum,0,sizeof(sum));
for(int j=f[i].h;j<=f[i].a;j++){
if(!dp[j]&&dp[j-f[i].h]&&sum[j-f[i].h]<f[i].c){
dp[j]=1;
sum[j]=sum[j-f[i].h]+1;
if(j>ans)ans=j;
}
}
}
printf("%d\n",ans);
return 0;
}