POJ - 2392 Space Elevator(多重背包）

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Submit Status

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.

注意背包容量

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
using namespace std;
typedef long long LL;
const int N = 201000;
int dp[N];
struct node
{
    int h,num, a;
}p[N];
int cmp(node x, node y)
{
    return x.a<y.a;
}


int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d %d %d", &p[i].h, &p[i].a,&p[i].num );
        }
        sort(p,p+n,cmp);
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=p[i].a;j++)
            {
                if(dp[j]>=0)
                {
                    dp[j]=p[i].num;
                    ans=max(j,ans);
                }
                else if(j<p[i].h||dp[j-p[i].h]<=0)
                {
                    dp[j]=-1;
                }
                else
                {
                    dp[j]=dp[j-p[i].h]-1;
                    ans=max(j,ans);
                }
            }
        }
        printf("%d\n",ans);
    }


    return 0;
}