PKU&EOJ：Space Elevator

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3691		Accepted: 1642

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

37 40 35 23 82 52 6

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.

原题地址：http://acm.pku.edu.cn/JudgeOnline/problem?id=2392

http://www.cn210.com/onlinejudge/problemshow.php?pro_id=10

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题解：

显然是背包问题。

————

可是我一直纠结在背包的容量到底是什么上。我的想法是，背包容量是物品的个数，然后背包的价值就是能达到的最大高度，即DP[I][J]表示前I个物品，一共用了J个物品，能达到的最大高度。后来发现这个状态表示的方法显然是有问题的，文中有限制条件，第I种物品，它的高度适中不能超过A[I]，那么就会发生这种情况：DP[I-1][J]+H[I]>A[I]，那么这个高度就达不到，但实际上前I-1种物品选了J个，有一种情况恰能达到A[I]-H[I]的高度，这样再选第I种物品，A[I]的高度就能达到了。可是A[I]-H[I]这个高度不是DP[I-1][J]里面的最大值，所以不会被记录下来。

————

正解：DP[I]表示高度为I能不能达到，实际上这题中，背包的容量和价值是同一个东西，即高度。这是一个多重背包问题，有O(N*V*C)的解法和O(V*Σlog C[i])的解法。另外，EOJ的评测机果然比PKU的慢很多。

O(N*V*C)

代码

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<memory.h>
 4  struct node
 5 {
 6     int h,a;
 7 }d[4001];
 8  int i,j,max,t,x,y,z,n;
 9  int dp[40001];
10  int cmp(const void *a,const void *b)
11 {
12     struct node *x=(struct node *)a;
13     struct node *y=(struct node *)b;
14     return ((x->a>y->a)?1:-1);
15 }
16  int main()
17 {
18     scanf("%d",&n);
19     t=0;
20     for (i=1;i<=n;i++)
21     {
22         scanf("%d%d%d",&x,&y,&z);
23         for (j=1;j<=z;j++)
24         {
25             d[++t].h=x;
26             d[t].a=y;
27         }
28     }
29     qsort(&d[1],t,sizeof(d[1]),cmp);
30     memset(dp,0,sizeof(dp));
31     dp[0]=1;
32     max=0;
33     for (i=1;i<=t;i++)
34         for (j=max;j>=0;j--)
35             if (dp[j]==1&&j+d[i].h<=d[i].a)
36             {
37                 dp[j+d[i].h]=1;
38                 if (j+d[i].h>max) max=j+d[i].h;
39             }
40     printf("%d\n",max);
41     return 0;
42 }
43

O(V*Σlog C[i])

代码

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<memory.h>
 4  struct node
 5 {
 6     int h,a;
 7 }d[4001];
 8  int i,j,max,t,x,y,z,n,rest,base;
 9  int dp[40001];
10  int cmp(const void *a,const void *b)
11 {
12     struct node *x=(struct node *)a;
13     struct node *y=(struct node *)b;
14     return ((x->a>y->a)?1:-1);
15 }
16  int main()
17 {
18     scanf("%d",&n);
19     t=0;
20     for (i=1;i<=n;i++)
21     {
22         scanf("%d%d%d",&x,&y,&z);
23         rest=z;
24         base=1;
25         while (base<=rest)
26         {
27             d[++t].h=x*base;
28             d[t].a=y;
29             rest-=base;
30             base*=2;
31         }
32         if (rest>0)
33         {
34             d[++t].h=x*rest;
35             d[t].a=y;
36         }
37     }
38     qsort(&d[1],t,sizeof(d[1]),cmp);
39     memset(dp,0,sizeof(dp));
40     dp[0]=1;
41     max=0;
42     for (i=1;i<=t;i++)
43         for (j=max;j>=0;j--)
44             if (dp[j]==1&&j+d[i].h<=d[i].a)
45             {
46                 dp[j+d[i].h]=1;
47                 if (j+d[i].h>max) max=j+d[i].h;
48             }
49     printf("%d\n",max);
50     return 0;
51 }
52

posted on 2010-07-29 14:25 风也轻云也淡阅读(201) 评论(0) 编辑收藏举报