POJ_2392_SpaceElevator（多重背包）

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10883		Accepted: 5185

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

给出了一些砖块

砖块有高度，最高可以达到的高度和数量

问最高可以摞多高的塔

解题思路

按照砖块最大高度排序

因为最大高度小的砖块先选肯定优于后选（贪心）

随后对于每个砖块做多重背包就可以了

注意的就是最后最大高度的dp[mh]不一定是最优解

因为这个dp[mh]的最值是从上一步过来的

而由于上一步dp的mh更小，因此dp数组存在断档

要最后从头扫一遍才能得出最优解

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int MV=4e4+5;
const int MN=40005;
int dp[MV];
int vm;
struct Block
{
    int h,mh,n;
    bool operator<(Block a)const
    {
        return mh<a.mh;
    }
}block[MN];
void _zop(int vi,int wi)//01背包子过程
{
    for(int i=vm;i>=vi;i--)
        dp[i]=max(dp[i],dp[i-vi]+wi);
}
void _cp(int vi,int wi)//完全背包子过程
{
    for(int i=vi;i<=vm;i++)
        dp[i]=max(dp[i],dp[i-vi]+wi);
}
void _mulp(int vi,int wi,int ni)//多重背包子过程
{
    if(wi*ni>=vm)
    {
        _cp(vi,wi);
        return;
    }
    int k=1;
    while(k<ni)
    {
        _zop(vi*k,wi*k);
        ni-=k;
        k<<=1;
    }
    _zop(vi*ni,wi*ni);
}

int main()
{
    int k;
    scanf("%d",&k);
    for(int i=0;i<k;i++)
        scanf("%d%d%d",&block[i].h,&block[i].mh,&block[i].n);
    sort(block,block+k);
    memset(dp,0,sizeof(dp));
    int ans=0;
    for(int i=0;i<k;i++)
    {
        vm=block[i].mh;
        _mulp(block[i].h,block[i].h,block[i].n);
        //ans=max(ans,dp[vm]);
    }
    for(int i=0;i<=vm;i++)
        ans=max(ans,dp[i]);
    printf("%d\n",ans);
    return 0;
}