poj2392 Space Elevator 多重背包转换为完全背包

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

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.

刚开始没反应过来，后来看了别人的才知道，刚开始一直在想最大值是多少，很巧妙的就是最高的呢个的限制的值，

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{    int h,lim,num;}q[500];int cmp(node a,node b){    return a.lim<b.lim;}int f[400000],v[400000];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)            scanf("%d %d %d",&q[i].h,&q[i].lim,&q[i].num);        sort(q,q+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 = q[i].h; j<=q[i].lim; j++)            {                if(!f[j] && f[j-q[i].h] && v[j-q[i].h]<q[i].num)                {                    f[j] = 1;                    v[j] = v[j-q[i].h]+1;                    if(ans<j)                        ans = j;                }            }        }        printf("%d\n",ans);    }    return 0;}