poj2392 Space Elevator（多重背包问题）

Space Elevator

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8569		Accepted: 4052

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.

 

题目大意：一群牛要上太空，给出n种石块，每种石块给出单块高度，每种石块都有个限定高度，超过这个高度这种石块就不能被使用了，最后面是每种石块的数量，要求用这些石块能组成的最大高度，并且不能超过限定的高度；

思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包了

 AC代码(一): PZ

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
#define max 40005
int a[max],b[max],c[max];
struct pp
{
    int x,y,z;
}p[405];
int cmp(pp p1,pp p2)//按照限定高度进行排序
{
    return p1.y<p2.y;
}
int main()
{
    int n;
    while((scanf("%d",&n))!=EOF)
    {
        int i,j,k,t;
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
            scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].z);
        sort(p,p+n,cmp);
        j=0;
        //暴力枚举所有的高度，a[s]=1; 表示s这个高度能够达到；
        for(i=0;i<n;i++)
        {
            int s=0,t=1,dd=0;
            while(s<=p[i].y&&t<=p[i].z) //小于限定的高度，并且小于限定的个数
            {
                s=s+p[i].x;
                for(k=0;k<j;k++) 
                    if(!a[b[k]+s] && (b[k]+s<=p[i].y)) //如果b[k]+s  这个高度没有出现过，并且小于限定的高低，高度可行
                    {
                        a[b[k]+s]=1;
                        c[dd++]=b[k]+s;
                    }
                if(!a[s] && s<=p[i].y) 
                {
                   a[s]=1;
                   b[j++]=s;
                }
                t++;              //统计当前石块使用个数
            }
            for(k=0;k<dd;k++)
                b[j++]=c[k];
        }
        int ok=1;
        for(i=40000;i>=0;i--)
            if(a[i])
            {
                printf("%d
",i);
                ok=0;
                break;
            }
        if(ok)
            printf("0
");
    }
    return 0;
}

 AC代码(二)：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 400010;
struct TT
{
    int x,h,n;
}a[410];
int dp[N],cot[N];
int cmp(TT x,TT y)//按照最高能堆多高进行排序；
{
    if( x.h<y.h)  return 1;
    return 0;
}
int main()
{
    int T,ans;
    while(cin>>T)
    {
        for(int i=0; i<T; i++)
            scanf("%d %d %d",&a[i].x,&a[i].h,&a[i].n);
            memset(dp,0,sizeof(0));
            sort(a,a+T,cmp);
            dp[0]  = 1;
            ans = 0;
            //采用暴力的方法，一直枚举j，看j最大能达到多少，则j就是要找的最大值；
            for(int i=0; i<T; i++)
            {
                memset(cot,0,sizeof(cot));
                for(int j = a[i].x; j<=a[i].h; j++)
                {
                // 如果j还没有达到并且dp[j-a[i].x] 大于0，并且当前模板的使用数不大于它的总数
                    if(!dp[j] && dp[j-a[i].x] && cot[j-a[i].x] < a[i].n) 
                    {
                        dp[j] = 1;
                        cot[j]=cot[j-a[i].x]+1;
                        if(j>ans) ans = j;
                    }
                }
            }
       printf("%d
",ans);
    }
    return 0;
}