POJ 2392——Space Elevator【多重背包 & 二进制优化 & 贪心】

题目传送门

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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.

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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

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Sample Input

3
7 40 3
5 23 8
2 52 6

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Sample Output

48

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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.

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题意：

n件物品，第i件hi高，有ci件，最高的一件不能超过ai的高度。问最高能堆多高

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分析：

• 多重背包 & 二进制优化
• 拆分时注意拆分上线并非为c，而应该是min(c, a/h)
• dp时注意添加判断条件

• 每次背包选取物品时都应该优先放置当前ai值最小的（贪心）

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AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define INF 0x3f3f3f

const int MAXN = 2e6 + 7;

int dp[MAXN];
struct node {
	int v, w;
	bool operator < (const node& t)const {
		return w < t.w;
	}
}node[MAXN];
int main() {
	ios;
	int n;
	while (cin >> n) {
		int sum = 0;
		memset(dp, 0, sizeof(dp));
		int k = 0;
		for (int i = 1; i <= n; i++) {
			int v, w, c;
			cin >> v >> w >> c;
			sum += c * v;
			c = min(c, w / v);
			for (int j = 1; j <= c; j *= 2) {
				node[k].v = v * j;
				node[k++].w = w;
				c -= j;
			}
			if (c > 0) {
				node[k].v = v * c;
				node[k++].w = w;
			}
		}
		sort(node, node + k);

		for (int i = 0; i < k; i++) {
			for (int j = sum; j >= node[i].v; j--)
				if (j <= node[i].w)
					dp[j] = max(dp[j], dp[j - node[i].v] + node[i].v);
		}
		int ans = 0;
		for (int i = sum; i >= 0; i--)
			ans = max(ans, dp[i]);
		cout << ans << endl;
	}
	return 0;
}