POJ-3624 (01背包)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 15677 | Accepted: 7130 |
Description
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
Output
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6 1 4 2 6 3 12 2 7
Sample Output
23
又领悟一个新的算法,01背包,也是作为动态规划的一个分支。..总结一下01背包的精髓就是通过数组来将背包容纳量单元化,通过嵌套的循环对每个单元格找出每个单元(即当前容量)的最优解法,循环过后即可得到最优解。
#include <iostream> #include <cstring> using namespace std; int max( int x, int y) { if (x>y) return x; return y; } int main() { int f[13000]; memset (f,0, sizeof f); int n,m; int w,d; cin>>n>>m; int i,j; for (i=1;i<=n;i++) { cin>>w>>d; for (j=m;j>=w;j--) { f[j]=max(f[j],f[j-w]+d); } } cout<<f[m]; } |
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