POJ 2392 Space Elevator （多重背包）

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6422		Accepted: 2965

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

 

 

先按照限制的高度进行从小到大排序。之后就是简单的多重背包了。

 

/*

POJ 2392

G++ 157ms

*/

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<algorithm>

using namespace std;

const int MAXN=410;



struct Node

{

    int h,c;

    int lim;

}node[MAXN];

int dp[50000];

int H;

void ZeroOnePack(int cost,int weight,int lim)

{

    for(int i=lim;i>=cost;i--)

       dp[i]=max(dp[i],dp[i-cost]+weight);

}



void CompletePack(int cost,int weight,int lim)

{

    for(int i=cost;i<=lim;i++)

      dp[i]=max(dp[i],dp[i-cost]+weight);

}



void MultiplePack(int cost,int weight,int amount,int lim)

{

    if(cost*amount>=lim)CompletePack(cost,weight,lim);

    else

    {

        for(int k=1;k<amount;)

        {

            ZeroOnePack(k*cost,k*weight,lim);

            amount-=k;

            k<<=1;

        }

        ZeroOnePack(amount*cost,amount*weight,lim);

    }

}



bool cmp(Node a,Node b)

{

    return a.lim<b.lim;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;



    while(scanf("%d",&n)!=EOF)

    {

        H=0;

        for(int i=0;i<n;i++)

        {

            scanf("%d%d%d",&node[i].h,&node[i].lim,&node[i].c);

            if(node[i].lim>H)H=node[i].lim;

        }

        for(int i=0;i<=H;i++)dp[i]=0;

        sort(node,node+n,cmp);

        for(int i=0;i<n;i++)

          MultiplePack(node[i].h,node[i].h,node[i].c,node[i].lim);

        int ans=0;

        for(int i=0;i<=H;i++)ans=max(ans,dp[i]);//这个过程一定要

        printf("%d\n",ans);

    }

    return 0;

}

/*

POJ 2392

G++ 141ms

*/

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<algorithm>

using namespace std;

const int MAXN=410;



struct Node

{

    int h,c;

    int lim;

}node[MAXN];

bool dp[50000];

int H;

int ans;

void ZeroOnePack(int cost,int lim)

{

    for(int i=lim;i>=cost;i--)

       if(dp[i-cost]&&(!dp[i]))

       {

           dp[i]=true;

           ans=max(ans,i);

       }

}



void CompletePack(int cost,int lim)

{

    for(int i=cost;i<=lim;i++)

      if(dp[i-cost]&&(!dp[i]))

      {

          dp[i]=true;

          ans=max(ans,i);

      }

}



void MultiplePack(int cost,int amount,int lim)

{

    if(cost*amount>=lim)CompletePack(cost,lim);

    else

    {

        for(int k=1;k<amount;)

        {

            ZeroOnePack(k*cost,lim);

            amount-=k;

            k<<=1;

        }

        ZeroOnePack(amount*cost,lim);

    }

}



bool cmp(Node a,Node b)

{

    return a.lim<b.lim;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;



    while(scanf("%d",&n)!=EOF)

    {

        H=0;

        for(int i=0;i<n;i++)

        {

            scanf("%d%d%d",&node[i].h,&node[i].lim,&node[i].c);

            if(node[i].lim>H)H=node[i].lim;

        }

        for(int i=0;i<=H;i++)dp[i]=false;

        dp[0]=true;

        sort(node,node+n,cmp);

        ans=0;

        for(int i=0;i<n;i++)

          MultiplePack(node[i].h,node[i].c,node[i].lim);

        printf("%d\n",ans);

    }

    return 0;

}