POJ 2392 Space Elevator

Space Elevator

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 3

Problem 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 <br> <br>* 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

 

题意:有一群牛要上太空,他们计划建一个太空梯(用一些石头垒),他们有k种不同类型的石头,每一种石头的高度为h,数量为c,由于会受到太空辐射,每一种石头不能超过这种石头的最大建造高度a,求解利用这些石头所能修建的太空梯的最高的高度.
    多重背包问题,与一般的多重背包问题所不同的知识多了一个限制条件就是某些"物品"叠加起来的"高度"不能超过一个值,于是我们可以对他们的最高可能达到高度进行排序,然后就是一般的多重背包问题了

 

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;
int dp[400005];
int v[40005];
struct node
{
    int h,a,c;
}g[40005];
bool cmp(node a, node b)
{
    return a.a < b.a;
}
int main()
{
    int k;
    cin >> k;
    int j, i;
    for (i = 1; i <= k; i++)
    {
        cin >> g[i].h >> g[i].a >> g[i].c;
    }
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    sort(g + 1, g + k + 1, cmp);
    int ans = 0;
    for (i = 1; i <= k; i++)
    {
        memset(v, 0, sizeof(v));//记录当前用了几个g[i]
        for (j = g[i].h; j <= g[i].a; j++)
        {
            if (!dp[j] && dp[j - g[i].h] && v[j - g[i].h] + 1 <= g[i].c)
            {//dp用来记录能不能达到着高度
                dp[j] = 1;
                v[j] = v[j - g[i].h] + 1;
                if (ans < j)
                    ans = j;
            }
        }
    }
    cout << ans << endl;
    return 0;
}