POJ 2392 Space Elevator(多重背包)

http://poj.org/problem?id=2392

 

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


题意:有一头奶牛要上太空,他有很多种石头,每种石头的高度是hi,
但是不能放到ai之上的高度,并且这种石头有ci个
将这些石头叠加起来,问能够达到的最高高度。




把多重背包转化为01背包
#include<cstdio>//把多重背包转化为01背包
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct DP
{
    int h,a,c;
};
DP num[400+10];
int dp[40000+100];
int cmp(DP x,DP y)
{
   return  x.a<y.a;
}
int main()
{
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&num[i].h,&num[i].a,&num[i].c);
        }
        sort(num+1,num+n+1,cmp);

        for(i=1;i<=n;i++)
        {
            for(j=1;j<=num[i].c;j++)
            {
                for(k=num[i].a;k>=0;k--)
                {
                    if(dp[k]==1&&k+num[i].h<=num[i].a)
                    {
                        dp[k+num[i].h]=1;
                    }
                }
            }
        }
        for(i=40000;i>=0;i--)
        {
            if(dp[i]==1) break;
        }
        printf("%d
",i);
    }
    return 0;
}

  

二进制优化

#include <cstring>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <iostream>
using namespace std;
int n,dp[45000];
struct node
{
    int a,b,c;
}e[500];
bool cmp(node a, node b)
{
    return a.b < b.b;
}
int main()
{
    int i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(i = 0; i < n ; i++)
        {
            scanf("%d %d %d",&e[i].a,&e[i].b,&e[i].c);
        }
        memset(dp, -1 , sizeof(dp));
        sort(e,e + n, cmp);
        dp[0] = 1;
        for(i = 0; i < n; i++)
        {
            int sum = 1;
            int q;
            for( k = 1; k<=e[i].c; k *= 2)
            {
                q = k * e[i].a;
                for(j = e[i].b; j >=0; j --)
                {
                    if(dp[j] != -1 && j + q <= e[i].b)
                    {
                        dp[j + q] = 1;
                    }
                }
                e[i].c -= k;
            }
            q = e[i].c * e[i].a;
            for(j = e[i].b; j >=0; j --)
            {
                if(dp[j] != -1 && j + q <= e[i].b)
                {
                    dp[j + q] = 1;
                }
            }
        }
        for(i = 40000; i >= 0; i--)
        {
            if(dp[i] == 1)
                break;
        }
        cout<<i<<endl;
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/sola1994/p/4090444.html