POJ_2392_Space_Elevator_(动态规划,背包)

描述

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http://poj.org/problem?id=2392

磊方块,每种方块有数量,高度,以及该种方块所能处在的最高高度.问最高磊多高?

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10511		Accepted: 4997

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.

Source

USACO 2005 March Gold

分析

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看起来很像多重背包问题,只不过这里的重量和价值是一样的,并且多了一个所能处在的最高高度的限制.

考虑这个限制,设i种所能处在的最高高度为hi.对于两种方块i,j,设hi<hj.那么放的时候肯定是先放i比较好,因为如果先放了j的话,i只能放在j的上面,对于一种合法的情况,我们把在下面的h较大的,和在上面的h较小的调换位置是可以的,但是把在下面的h较小的和在上面的h较大的调换位置却不一定可以,所以应该尽量避免h小的放在上面,所以要先放i.放第i种方块的时候背包的容量上限是hi.

注意:

1.数组大小,好几次都没开够,没有好好看数据范围啊...

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #define for1(i,a,n) for(int i=(a);i<=(n);i++)
 #define read(a) a=getnum()
 using namespace std;

 const int maxn=+,maxh=+;
 int n;
 int dp[maxh];
 struct node { int h,v,m; }a[maxn];

 inline int getnum(){int r=;char c;c=getchar();while(c<''||c>'')c=getchar();while(c>=''&&c<=''){r=r*+c-'';c=getchar();}return r;}

 bool comp(node x,node y) { return x.h<y.h; }

 void solve()
 {
     sort(a+,a+n+,comp);
     int ans=;
     for1(i,,n)
     {
         for(int k=;k<a[i].m;k*=)
         {
             for(int j=a[i].h;j>=k*a[i].v;j--)
             {
                 dp[j]=max(dp[j],dp[j-k*a[i].v]+k*a[i].v);
                 ans=max(ans,dp[j]);
             }
             a[i].m-=k;
         }
         for(int j=a[i].h;j>=a[i].m*a[i].v;j--)
         {
             dp[j]=max(dp[j],dp[j-a[i].m*a[i].v]+a[i].m*a[i].v);
             ans=max(ans,dp[j]);
         }
     }
     printf("%d\n",ans);
 }

 void init()
 {
     read(n);
     for1(i,,n)
     {
         read(a[i].v);
         read(a[i].h);
         read(a[i].m);
     }
 }

 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("space.in","r",stdin);
     freopen("space.out","w",stdout);
 #endif
     init();
     solve();
 #ifndef ONLINE_JUDGE
     fclose(stdin);
     fclose(stdout);
     system("space.out");
 #endif
     return ;
 }