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//把多重背包转化为01背包
#include
#include
#include
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=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\n",i);
    }
    return 0;
}

　　

 

二进制优化

#include 
#include 
#include 
#include 
#include 
#include 
#include 
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<

　　

转载于:https://www.cnblogs.com/sola1994/p/4090444.html