poj 2392 Space Elevator(多重背包变形)

Space Elevator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5144   Accepted: 2382 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

题目：http://poj.org/problem?id=2392

分析：这题其实就是多重背包加一个排序，先按最高能达到的海拔排序，然后直接多重背包= =

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=444;
struct block
{
    int h,a,c;
}g[mm];
bool f[44444];
int i,j,k,n;
bool cmp(block a,block b)
{
    return a.a<b.a;
}
int main()
{
    while(scanf("%d",&n)!=-1)
    {
        for(i=0;i<n;++i)scanf("%d%d%d",&g[i].h,&g[i].a,&g[i].c);
        sort(g,g+n,cmp);
        for(f[0]=1,i=g[n-1].a;i>0;--i)f[i]=0;
        for(i=0;i<n;++i)
            for(j=0;j<g[i].c;++j)
                for(k=g[i].a;k>=g[i].h;--k)
                    f[k]|=f[k-g[i].h];
        for(i=g[n-1].a;i>0;--i)
            if(f[i])break;
        printf("%d\n",i);
    }
    return 0;
}