poj 3624 01背包

Posted 2021-02-13 lxc1910

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 W_i (1 ≤ W_i ≤ 400), a ‘desirability‘ factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

Source

USACO 2007 December Silver

dp[i][j]代表当背包体积为j时，前i件物品可以达到的最大价值。

当i为0时，前0件物品价值只能是0。所以第一行全是0；当第i件物品重量大于j时，dp[i][j] = dp[i - 1][j]，否则dp[i][j] = max(dp[i - 1][j],

dp[i - 1][j - w[i] ] + v[i])。

为了优化内存，可以使用一维数组，但是j要从M递减到1，否则从1到M的话前面的会覆盖掉后面要用到的值。

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 vector<int> dp;
 7 
 8 int main() {
 9     int N, M, i, j;
10     cin >> N >> M;
11     vector<int> W(N + 1), D(N + 1);
12     for (i = 1; i <= N; i++)
13         cin >> W[i] >> D[i];
14     dp.resize(M + 1);
15     for (i = 1; i <= N; i++) {
16         for (j = M; j >= 1; j--) {
17             if (W[i] <= j)
18                 dp[j] = max(dp[j], dp[j - W[i]] + D[i]);
19         }
20     }
21     cout << dp[M] << endl;
22     return 0;
23 }