POJ2392Space Elevator(贪心+背包）

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9970		Accepted: 4738

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.

题意：奶牛想上太空给顶n种梯子，每种梯子对应三个值，a,h,c，a表示这种梯子必须在小于等于a的高度内使用，h表示它的高度，c表示这种梯子的个数。问内牛能够累出的最大高度。

分析：根据a从小到大排序，然后对于每一个找到背包容量为a的最大高度，可以用01背包处理，对每一个奶牛的梯子作为一个物品，物品的种数就是c;当然也可以用多重背包解

思路就是辣么简单就是没想出来，弱渣

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int MAX = 50000;
int dp[MAX];
struct node
{
    int h,a,c;
};
node cow[500];
int cmp(node x, node y)
{
    return x.a < y.a;
}

int main()
{
    int k;
    while(scanf("%d", &k) != EOF)
    {
        for(int i = 0; i < k; i++)
        {
            scanf("%d%d%d",&cow[i].h,&cow[i].a,&cow[i].c);
        }
        memset(dp,0,sizeof(dp));
        sort(cow, cow + k, cmp);
        for(int i = 0; i < k; i++)
        {
            for(int t = 1; t <= cow[i].c; t++)
            {
                for(int j = cow[i].a; j >= cow[i].h; j--)
                {
                    if(dp[j] <= cow[i].a && dp[j - cow[i].h] + cow[i].h <= cow[i].a)
                    dp[j] = max(dp[j], dp[j - cow[i].h] + cow[i].h);
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= cow[k - 1].a; i++)  //这一步还是在斌神那里得到的提示，太弱了
            ans = max(ans, dp[i]);
        printf("%d
", ans);
    }
    return 0;
}