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*，每种*对应三个值，a,h,c，a表示这种*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*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*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 = ;
int dp[MAX];
struct node
{
int h,a,c;
};
node cow[];
int cmp(node x, node y)
{
return x.a < y.a;
} int main()
{
int k;
while(scanf("%d", &k) != EOF)
{
for(int i = ; i < k; i++)
{
scanf("%d%d%d",&cow[i].h,&cow[i].a,&cow[i].c);
}
memset(dp,,sizeof(dp));
sort(cow, cow + k, cmp);
for(int i = ; i < k; i++)
{
for(int t = ; 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 = ;
for(int i = ; i <= cow[k - ].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\n", ans);
}
return ;
}