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[];
 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,,sizeof(a));
         for(i=;i<n;i++)
             scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].z);
         sort(p,p+n,cmp);
         j=;
         //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=;i<n;i++)
         {
             int s=,t=,dd=;
             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=;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]=;
                         c[dd++]=b[k]+s;
                     }
                 if(!a[s] && s<=p[i].y)
                 {
                    a[s]=;
                    b[j++]=s;
                 }
                 t++;              //统计当前石块使用个数
             }
             for(k=;k<dd;k++)
                 b[j++]=c[k];
         }
         int ok=;
         for(i=;i>=;i--)
             if(a[i])
             {
                 printf("%d\n",i);
                 ok=;
                 break;
             }
         if(ok)
             printf("0\n");
     }
     return ;
 }

AC代码(二)：

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 const int N = ;
 struct TT
 {
     int x,h,n;
 }a[];
 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 ;
     return ;
 }
 int main()
 {
     int T,ans;
     while(cin>>T)
     {
         for(int i=; i<T; i++)
             scanf("%d %d %d",&a[i].x,&a[i].h,&a[i].n);
             memset(dp,,sizeof());
             sort(a,a+T,cmp);
             dp[]  = ;
             ans = ;
             //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=; i<T; i++)
             {
                 memset(cot,,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] = ;
                         cot[j]=cot[j-a[i].x]+;
                         if(j>ans) ans = j;
                     }
                 }
             }
        printf("%d\n",ans);
     }
     return ;
 }