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
题意:有一群牛要上太空,他们计划建一个太空梯(用一些石头垒),他们有k种不同类型的石头,每一种石头的高度为h,数量为c,由于会受到太空辐射,每一种石头不能超过这种石头的最大建造高度a,求解利用这些石头所能修建的太空梯的最高的高度.
多重背包问题,与一般的多重背包问题所不同的知识多了一个限制条件就是某些"物品"叠加起来的"高度"不能超过一个值,于是我们可以对他们的最高可能达到高度进行排序,然后就是一般的多重背包问题了
思路：
首先是录入数据，这一点比较简单，但是最好用scanf，因为它比cin快；
录入之后对各个数据按照最大的使用高度进行排序，就会转化成一个一般的多重背包为题，一开始我是用原始的多重背包做的麻烦并且也没有A了，最后看了一下题解，是因为每种块的使用个数的计算处理不当；
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<memory.h> 7 using namespace std; 8 int a[40005]; 9 struct node 10 { 11 int h; 12 int max; 13 int shu; 14 }; 15 node num[40005]; 16 int dp[40005]; 17 bool cmp(node a,node b) 18 { 19 return a.max<b.max; 20 } 21 int main() 22 { 23 // freopen("1.txt","r",stdin); 24 int i,j,k; 25 int m; 26 memset(dp,0,sizeof(dp)); 27 dp[0]=1; 28 int sum[40005]; 29 cin>>k; 30 for(i=0;i<k;i++) 31 { 32 scanf("%d%d%d",&num[i].h,&num[i].max,&num[i].shu); 33 } 34 sort(num,num+k,cmp); 35 m=num[k-1].max; 36 int ans=0; 37 for(i=0;i<k;i++) 38 { 39 memset(sum,0,sizeof(sum)); 40 for(j=num[i].h;j<=num[i].max;j++) 41 { 42 if(!dp[j]&&dp[j-num[i].h]&&sum[j-num[i].h]<num[i].shu) 43 { 44 dp[j]=1; 45 sum[j]=sum[j-num[i].h]+1;//提柜条件，表示到达j高度已经用的砖的个数 46 if(ans<j) 47 ans=j; 48 } 49 } 50 } 51 cout<<ans<<endl; 52 return 0; 53 }