1635 Space Elevator【解题报告】

 

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2829 Accepted: 1172

http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1635

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

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：

高度相当于物品的容量也是价值，而这里的限制高度相当于背包容量，当然这里有多只背包，但最后是组合起来的，相当于一只。运用多重背包，其实也是01背包的拓展而已，基本一样。

要想知道之间的关系参看：http://blog.csdn.net/helihui123/archive/2009/11/13/4807977.aspx

具体的都在那个参看那里，这里不多说直接代码，我加了注释：

#include<stdio.h>#include<stdlib.h>#include<string.h>#define max(x,y) x>y?x:yint f[40005];typedef struct{int a,h,c;}node;int cmp(void const *a,void const *b){ return (*(node *)a).a>(*(node *)b).a?1:-1;}void zeroonepack(int c,int cost,int weight)//01背包 c表示当前背包总容量，该物品的价值，与容量{int i;for(i=c;i>=weight;i--){f[i]=max(f[i-weight]+cost,f[i]);}}void completepack(int c,int cost,int weight)//完全背包{int i;if(weight<=c){for(i=weight;i<=c;i++){f[i]=max(f[i-weight]+cost,f[i]);}}}void mulitpack(int c,int cost,int weight,int count)//多重背包{int p;if(count*weight>=c){completepack(c,weight,weight);}else{p=1;while(p<count){zeroonepack(c,p*weight,p*weight);count-=p;p*=2;}p=count;zeroonepack(c,p*weight,p*weight);}}int main(){node wu[405];int k,i,mmax;while(scanf("%d",&k)!=EOF){mmax=0;memset(f,0,sizeof(f));for(i=0;i<k;i++){scanf("%d%d%d",&wu[i].h,&wu[i].a,&wu[i].c);}qsort(wu,k,sizeof(wu[0]),cmp);for(i=0;i<k;i++){mulitpack(wu[i].a,wu[i].h,wu[i].h,wu[i].c);}for(i=0;i<=wu[k-1].a;i++)//这里为什么要这样取最大值呢？看看数据就知道了，因为如果最大背包容量大于物品容量和 优化如何？自己想吧{printf("%d/n",f[i]);mmax=max(mmax,f[i]);}printf("%d/n",mmax);}return 0;}/*23 49 322 77 2这组数据就是说明了上面的问题的*/