POJ 2392 Space Elevator 多重背包
2014-02-18 16:34
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看见啸爷再看这道题目,也跟着看了看,看懂之后感觉有戏,很裸的背包,果然1A,嘿嘿。
题目大意:给你n种木块,然后让你输出最高可以组成的高度。
限制条件是:每种木块的个数,与木块的在高度h以上就不可以再出现了。
解题思路:根据每种木块可以到达的高度sort一遍然后就是多重背包,找到满足条件的最大的高度。
注意可以到达的最大的高度不会超过sort之后木块可以到达的上限,f[n-1].lim,因为在往上,是不能再搭木块的啊,所以就把这个高度当作,dp的上限,改了一下竟然省了300+ms。。。
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
题目大意:给你n种木块,然后让你输出最高可以组成的高度。
限制条件是:每种木块的个数,与木块的在高度h以上就不可以再出现了。
解题思路:根据每种木块可以到达的高度sort一遍然后就是多重背包,找到满足条件的最大的高度。
注意可以到达的最大的高度不会超过sort之后木块可以到达的上限,f[n-1].lim,因为在往上,是不能再搭木块的啊,所以就把这个高度当作,dp的上限,改了一下竟然省了300+ms。。。
Space Elevator
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7657 | Accepted: 3621 |
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
#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <map> #include <set> #define eps 1e-7 #define M 1000100 //#define LL __int64 #define LL long long #define INF 0x3f3f3f3f #define PI 3.1415926535898 const int maxn = 500100; using namespace std; int dp[maxn]; struct node { int h, num, lim; } f[maxn]; int cmp(node a, node b) { return a.lim < b.lim; } int main() { int n; cin >>n; for(int i = 0; i < n; i++) cin >>f[i].h>>f[i].lim>>f[i].num; sort(f , f+n, cmp); memset(dp , 0 , sizeof(dp)); dp[0] = 1; for(int i = 0; i < n; i++) { for(int j = f[n-1].lim; j >= 0; j--) { if(!dp[j]) continue; for(int k = 1; k <= f[i].num; k++) { if(j+k*f[i].h <= f[i].lim) { if(!dp[j+k*f[i].h]) dp[j+k*f[i].h] = 1; else break; } } } } int i; for(i = f[n-1].lim; i >= 0; i--) if(dp[i]) break; cout<<i<<endl; return 0; }
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