题目链接：https://ac.nowcoder.com/acm/contest/993/H/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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.

输入描述

* 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

输出描述

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

输入

4 6
1 4
2 6
3 12
2 7

输出

23

解题思路

题意:在最多m权重的情况下求出最大的魅力和。
思路:01背包。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 12885;
int dp[MAXN];
int main() {
    int n, m, w, v;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &w, &v);
        for (int j = m; j >= w; j--)
            dp[j] = max(dp[j], dp[j - w] + v);
    }
    printf("%d\n", dp[m]);
    return 0;
}