POJ 2392 Space Elevator

Space Elevator

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 11622 Accepted: 5518

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

思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包

思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;
const int maxn = 400000;
struct node
{
int h_i;
int a_i;
int c_i;
bool operator < (const node &n)const
{
return a_i < n.a_i;
}
}vis[405];

int dp[maxn
de31
];
int cnt[maxn];
int main()
{
int k;
while(cin>>k)
{
for(int i=0;i<k;i++)
{
cin>>vis[i].h_i>>vis[i].a_i>>vis[i].c_i;
}
sort(vis,vis+k);
memset(dp,0,sizeof(dp));
dp[0] = 1;
int ans = 0;
for(int i=0;i<k;i++)
{
memset(cnt,0,sizeof(cnt));
for(int j=vis[i].h_i;j<=vis[i].a_i;j++)
{
// 维持单调队列的递减特性
if(!dp[j]&&dp[j-vis[i].h_i]&&cnt[j-vis[i].h_i]<vis[i].c_i)
{
dp[j]=1;
cnt[j]=cnt[j-vis[i].h_i]+1;
if(ans<j)
ans=j;
}
}
}
cout<<ans<<endl;
}
return 0;
}