POJ2392-Space Elevator
2017-05-21 23:10
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Space Elevator
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
Sample Output
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.
Source
USACO 2005 March Gold
题意:一群牛要上太空,给出n种石块,每种石块给出单块高度,总高度不能超过的最大值,数量,用这些石块能组成的最大高度
解题思路:先进行一次排序,将最大高度小的放在前面,只有这样才能得到最优解,如果将大的放在前面,后面有的小的就不能取到,然后进行完全背包
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11818 | Accepted: 5601 |
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.
Source
USACO 2005 March Gold
题意:一群牛要上太空,给出n种石块,每种石块给出单块高度,总高度不能超过的最大值,数量,用这些石块能组成的最大高度
解题思路:先进行一次排序,将最大高度小的放在前面,只有这样才能得到最优解,如果将大的放在前面,后面有的小的就不能取到,然后进行完全背包
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <map> #include <cmath> #include <set> #include <stack> #include <queue> #include <vector> #include <bitset> #include <functional> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; struct node { int h, a, c; }x[409]; int cmp(node x, node y) { return x.a<y.a; } int dp[50090], sum[50090]; int main() { int n; while (~scanf("%d", &n)) { for (int i = 1; i <= n; i++) scanf("%d%d%d", &x[i].h, &x[i].a, &x[i].c); sort(x + 1, x + n + 1, cmp); memset(dp, 0, sizeof dp); dp[0] = 1; int ma = 0; for (int i = 1; i <= n; i++) { memset(sum, 0, sizeof sum); for (int j = x[i].h; j <= x[i].a; j++) { if (!dp[j] && dp[j - x[i].h] && sum[j - x[i].h] + 1 <= x[i].c) { dp[j] = 1; sum[j] = sum[j - x[i].h] + 1; ma=max(ma, j); } } } printf("%d\n", ma); } return 0; }
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