Charm Bracelet POJ - 3624 (利用滚动数组解决01背包)

雨点打透心脏的1/2处 2022-02-24 14:54 233阅读 0赞

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

解题思路:

01背包问题,由于数组太大会超内存,这里要使用滚动数组解决

#include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <string.h>
    using namespace std;
    int N, M;
    const int MAXN = 3500;  //最大容量
    const int MAXV = 13000;  //最大重量
    int W[MAXN];  //各物品重量
    int D[MAXN];  //各物品价值
    int sdp[MAXV];  //一维数组,用来滚动操作
    int main() {
    	cin >> N >> M;
    	for (int i = 1; i <= N; ++i) {  //读入数据
    		cin >> W[i] >> D[i];   //读入重量和价值
    	}
    	memset(sdp, 0, sizeof(sdp));  
    	for (int i = 1; i <= N; ++i) {
    		for (int j = M; j >= W[i]; --j) {  //从后往前推
    			sdp[j] = max(sdp[j], sdp[j - W[i]] + D[i]);
    		}
    	}
    	cout << sdp[M] << endl;
    	system("PAUSE");
    	return 0;
    }