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 (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

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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<map>
 6 #include<set>
 7 using namespace std;
 8 #define N 406
 9 #define M 40006
10 struct Node{
11     int h,a,c;
12 }block[N];
13 int dp[M*N];
14 bool cmp(Node a,Node b){
15     return a.a<b.a;
16 }
17 int main()
18 {
19     int n;
20     while(scanf("%d",&n)==1){
21         for(int i=0;i<n;i++){
22             scanf("%d%d%d",&block[i].h,&block[i].a,&block[i].c);
23         }
24         sort(block,block+n,cmp);
25
26        memset(dp,-1,sizeof(dp));
27        dp[0]=0;
28        for(int i=0;i<n;i++){
29            for(int j=0;j<=M;j++){
30                if(dp[j]>=0){
31                   dp[j]=block[i].c;
32                }
33                else if(j<block[i].h || dp[j-block[i].h]<=0){
34                   dp[j]=-1;
35                }
36                else if(j>block[i].a){
37                   dp[j]=-1;
38                }
39                else{
40                   dp[j]=dp[j-block[i].h]-1;
41                }
42            }
43        }
44        for(int i=M;i>=0;i--){
45           if(dp[i]!=-1){
46              printf("%d\n",i);
47              break;
48           }
49        }
50
51
52
53
54
55     }
56     return 0;
57 }