POJ 2392-Space Elevator(多重部分和-多重背包)

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10506		Accepted: 4994

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

Sample Output

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

题目意思：

有K组数，后面三个数h，a，c分别表示一块砖的高度、用这种砖的最大高度和用这种砖的最多块数。
求用这些砖叠在一起的最大高度。

解题思路：

用结构体根据“用这种砖的最大高度”升序排列，然后有两种思路：
①多重部分和；
②多重背包，限制高度作为背包中的最大高度。

两种方法解题占用内存一样，但是①比②快一倍。

AC代码①多重部分和

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 100010
int dp[MAXN];//d[i]表示石块能组成的高度为i
struct Node
{
    int h,a,c;//分别表示一块砖的高度、用这种砖的最大高度和最多块数
} s[MAXN];
int cmp(Node x,Node y)//结构体排序
{
    return x.a<y.a;//升序
}
int main()
{
    int n,i,j,ans=0;
    cin>>n;
    for(i=0; i<n; ++i)
        cin>>s[i].h>>s[i].a>>s[i].c;
    sort(s,s+n,cmp);//按最大高度升序排列
    memset(dp,-1,sizeof dp);
    dp[0]=0;
    for(i=0; i<n; i++)//多重部分和
        for(j=0; j<=s[i].a; j++)
        {
            if (dp[j]>=0) dp[j]=s[i].c;
            else if (j<s[i].h || dp[j-s[i].h] <=0) dp[j]=-1;
            else dp[j] = dp[j-s[i].h]-1;
        }
    for (i=s[n-1].a; i>=0; --i)
        if(dp[i]>=0)
        {
            ans=i;
            break;
        }
    cout<<ans<<endl;
    return 0;
}

AC代码②多重背包

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 100010//一开始开了10010一直RE…囧o(╯□╰)o
int dp[MAXN];//d[i]表示石块能组成的高度为i
struct Node
{
    int h,a,c;//分别表示一块砖的高度、用这种砖的最大高度和最多块数
} s[MAXN];
int cmp(Node x,Node y)//结构体排序
{
    return x.a<y.a;//升序
}
int main()
{
    int n,i,j,k,ans=0;
    cin>>n;
    for(i=0; i<n; ++i)
        cin>>s[i].h>>s[i].a>>s[i].c;
    sort(s,s+n,cmp);//按最大高度升序排列
    memset(dp,0,sizeof(dp));//初始化
    for(i=0; i<n; ++i)//n种砖块，使用多重背包
    {
        if(s[i].h*s[i].c>=s[i].a)//完全背包
            for(j=s[i].h; j<=s[i].a; j++)
                dp[j]=max(dp[j],dp[j-s[i].h]+s[i].h);
        else//01背包
        {
            for(k=1; k<=s[i].c; k<<=1) //数量
            {
                for(j=s[i].a; j>=k*s[i].h; --j)//高度
                    dp[j]=max(dp[j],dp[j-k*s[i].h]+k*s[i].h);
                s[i].c-=k;//减去用过的数量
            }
            for(j=s[i].a; j>=s[i].h*s[i].c; --j)
                dp[j]=max(dp[j],dp[j-s[i].h*s[i].c]+s[i].h*s[i].c);
        }
    }
    for(i=1; i<=s[n-1].a; ++i)//s[n-1].a表示所有限制高度中的最大值
        ans=max(ans,dp[i]);//找出石块能组成的最大高度
    cout<<ans<<endl;
    return 0;
}