• 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;
    }
    

      

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