poj 2392 Space Elevator(多重背包+先排序)

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 ( <= K <= ) different types of blocks with which to build the tower. Each block of type i has height h_i ( <= h_i <= ) and is available in quantity c_i ( <= c_i <= ). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a_i ( <= a_i <= ). 

Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.

Input

* Line : A single integer, K 

* Lines ..K+: Each line contains three space-separated integers: h_i, a_i, and c_i. Line i+ describes block type i.

Output

* Line : A single integer H, the maximum height of a tower that can be built

Sample Input

Sample Output

Hint

OUTPUT DETAILS: 

From the bottom:  blocks of type , below  of type , below  of type . Stacking  blocks of type  and  of type  is not legal, since the top of the last type  block would exceed height .

Source

USACO 2005 March Gold

 

这道题和 poj 1742 有点类似，只是这道题要先按每个block的最大高度进行排序，这样做的目的是为了获得最优解，想想怎么证明？

然后将dp数组初始化为-1，dp=-1表示取不到，dp[i]>=0表示取到i的时候还能剩下多少个

最后结果就是从最大高度开始寻找dp[i]!=-1的i的值，直接输出即可

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<map>
 #include<set>
 using namespace std;
 #define N 406
 #define M 40006
 struct Node{
     int h,a,c;
 }block[N];
 int dp[M*N];
 bool cmp(Node a,Node b){
     return a.a<b.a;
 }
 int main()
 {
     int n;
     while(scanf("%d",&n)==){
         for(int i=;i<n;i++){
             scanf("%d%d%d",&block[i].h,&block[i].a,&block[i].c);
         }
         sort(block,block+n,cmp);

        memset(dp,-,sizeof(dp));
        dp[]=;
        for(int i=;i<n;i++){
            for(int j=;j<=M;j++){
                if(dp[j]>=){
                   dp[j]=block[i].c;
                }
                else if(j<block[i].h || dp[j-block[i].h]<=){
                   dp[j]=-;
                }
                else if(j>block[i].a){
                   dp[j]=-;
                }
                else{
                   dp[j]=dp[j-block[i].h]-;
                }
            }
        }
        for(int i=M;i>=;i--){
           if(dp[i]!=-){
              printf("%d\n",i);
              break;
           }
        }

     }
     return ;
 }