Space Elevator
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13340 | Accepted: 6323 |
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.
我们先把石头达到的最大高度排序后，这个问题就转化成为了一个完全背包问题。
但是我们这次需要遍历一遍dp数组来求得最大值。（因为最终的高度不确定，即背包的容量就不确定）。
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn = 1000500;
int n,m;
struct node
{
int h,hh,num;
}s[maxn];
int dp[maxn];
bool cmp(node a,node b)
{
return a.hh<b.hh;
}
void zopac(int v,int h)//01背包
{
for(int j=v;j>=h;j--)
{
dp[j]=max(dp[j],dp[j-h]+h);
}
}
void copac(int v,int h)//完全背包
{
for(int j=h;j<=v;j++)
{
dp[j]=max(dp[j],dp[j-h]+h);
}
}
void mupac(int n)//多重背包
{
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
if(s[i].h*s[i].num>=s[i].hh) copac(s[i].hh,s[i].h);
else{
for(int k=1;k<=s[i].num;k<<=1)
{
zopac(s[i].hh,k*s[i].h);
s[i].num-=k;
}
if(s[i].num) zopac(s[i].hh,s[i].num*s[i].h);
}
}
}
int main()
{
cin>>n;
int cnt=0;
for(int i=0;i<n;i++){
cin>>s[i].h>>s[i].hh>>s[i].num;
cnt=max(cnt,s[i].hh);
}
sort(s,s+n,cmp);
mupac(n);
int ans=0;
for(int i=0;i<=cnt;i++) ans=max(ans,dp[i]);
cout<<ans<<endl;
return 0;
}