poj 2392 Space Elevator

Space Elevator

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.

 

这题首先要对每种石头能放的最大大高度进行排序，然后在进行取石头，用数组f[j]标记能到达的高度为j，同时还要use[i]数组记录到达高度j时每种石头用的个数。

代码如下

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct ss
{
    int h,a,c;
    /*bool operator<(const ss &g)const{
        return a<g.a;*/     //不用cmp进行排序而用注释那部分代码，效率更高
}num[410];
bool cmp(ss t1,ss t2)
{
    return t1.a<t2.a;
}
int f[50000],use[50000];
int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        int sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&num[i].h,&num[i].a,&num[i].c);
            sum=sum+num[i].c;
        }
        sort(num+1,num+n+1,cmp);
        memset(f,0,sizeof(f));
        f[0]=1;
        int max=0;
        for(i=1;i<=n;i++)
        {
            memset(use,0,sizeof(use));
            for(j=num[i].h;j<=num[i].a;j++)
            {
                if(!f[j]&&f[j-num[i].h]&&use[j-num[i].h]+1<=num[i].c)//如果f[j-num[i].h]！=0表示j-num[i].h这高度达到啦的
                {
                    f[j]=1;
                    use[j]=use[j-num[i].h]+1;
                    if(max<j) max=j; //记录达到的最大高度
                }
            }
        }
        printf("%d\n",max);
    }
    return 0;
}