POJ 2392 Space Elevator (多重背包问题)

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6714 Accepted: 3116

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

思路：因为每种类型的砖头都有一定的高度限制，这样只要根据每种砖头的高度限制进行升序排序，把限制当作背包容量，分别进行多重背包。

View Code

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define MAX 40005 6  7 using namespace std; 8  9 struct node10 {11     int height,limit,num;12 };13 14 node block[405];15 int dp[MAX];16 int K;17 18 bool cmp(node a,node b)19 {20     return a.limit < b.limit;21 }22 23 void multiPack(int ht,int n,int lmt)24 {25     if(ht * n > lmt)26     {27         for(int i=ht; i<=lmt; i++)28             dp[i] = max(dp[i],dp[i-ht]+ht);29     }30     else31     {32         int k = 1;33         int num = n;34         while(k < num)35         {36             for(int i=lmt; i>=ht*k; i--)37                 dp[i] = max(dp[i],dp[i-k*ht]+k*ht);38             num -= k;39             k *= 2;40         }41         for(int i=lmt; i>=ht*num; i--)42             dp[i] = max(dp[i],dp[i-num*ht]+num*ht);43     }44 }45 46 int main()47 {48     scanf("%d",&K);49     for(int i=1; i<=K; i++)50         scanf("%d%d%d",&block[i].height,&block[i].limit,&block[i].num);51     sort(block+1,block+K+1,cmp);52     memset(dp,0,sizeof(dp));53     for(int i=1; i<=K; i++)54         multiPack(block[i].height,block[i].num,block[i].limit);55     int ans = 0;56     for(int i=1; i<=block[K].limit; i++)57         if(ans < dp[i])58             ans = dp[i];59     printf("%d\n",ans);60     return 0;61 }