POJ_2392_Space_Elevator_(动态规划,背包)
描述
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.
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.
* 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.
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
分析
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...
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #define for1(i,a,n) for(int i=(a);i<=(n);i++) 5 #define read(a) a=getnum() 6 using namespace std; 7 8 const int maxn=400+5,maxh=40000+5; 9 int n; 10 int dp[maxh]; 11 struct node { int h,v,m; }a[maxn]; 12 13 inline int getnum(){int r=0;char c;c=getchar();while(c<'0'||c>'9')c=getchar();while(c>='0'&&c<='9'){r=r*10+c-'0';c=getchar();}return r;} 14 15 bool comp(node x,node y) { return x.h<y.h; } 16 17 void solve() 18 { 19 sort(a+1,a+n+1,comp); 20 int ans=0; 21 for1(i,1,n) 22 { 23 for(int k=1;k<a[i].m;k*=2) 24 { 25 for(int j=a[i].h;j>=k*a[i].v;j--) 26 { 27 dp[j]=max(dp[j],dp[j-k*a[i].v]+k*a[i].v); 28 ans=max(ans,dp[j]); 29 } 30 a[i].m-=k; 31 } 32 for(int j=a[i].h;j>=a[i].m*a[i].v;j--) 33 { 34 dp[j]=max(dp[j],dp[j-a[i].m*a[i].v]+a[i].m*a[i].v); 35 ans=max(ans,dp[j]); 36 } 37 } 38 printf("%d\n",ans); 39 } 40 41 void init() 42 { 43 read(n); 44 for1(i,1,n) 45 { 46 read(a[i].v); 47 read(a[i].h); 48 read(a[i].m); 49 } 50 } 51 52 int main() 53 { 54 #ifndef ONLINE_JUDGE 55 freopen("space.in","r",stdin); 56 freopen("space.out","w",stdout); 57 #endif 58 init(); 59 solve(); 60 #ifndef ONLINE_JUDGE 61 fclose(stdin); 62 fclose(stdout); 63 system("space.out"); 64 #endif 65 return 0; 66 }