Space Elevator

题目连接：https://cn.vjudge.net/contest/202335#problem/L

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. 

题意：通过用每种型号的模块构建最高的塔； 每种模块高度不同，数量有限，并且在空间中存在的高度有限。

多重背包转换为完全背包！！！

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int h,a,c;};int f[40010],v[40010];//f数组中存放几种类型模块可搭建的高度，v数组存放块数int cmp(node s,node b){    return s.a<b.a;}int main(){    int n;    scanf("%d",&n);    node num[n];    for(int i=0;i<n;i++)        scanf("%d %d %d",&num[i].h,&num[i].a,&num[i].c);    sort(num,num+n,cmp);    memset(f,0,sizeof(f));    f[0]=1;    int ans=0;    for(int i=0;i<n;i++)    {        memset(v,0,sizeof(v));        for(int j=num[i].h;j<=num[i].a;j++)        {            if(!f[j]&&f[j-num[i].h]&&v[j-num[i].h]<num[i].c)            {                f[j]=1;                v[j]=v[j-num[i].h]+1;                if(ans<j)                    ans=j;                //printf("j-num[%d].h=%d num[%d].h=%d\n",i,j-num[i].h,i,num[i].h);                //printf("ans=%d\n",ans);            }        }    }    printf("%d\n",ans);    return 0;}/*数组f中通过j++循环存放可能拼成的所有高度如果该种拼法高度存在，则f[i]=1,否则f[i]=0;通过if判断增加新拼法，进而更新高度v数组存放在每种限制（a）下的块数ans记录更新最高高度*/