有很多种写法，二维数组，一维数组等，这里只写滚动数组，这是优化最大的，而且往后完全背包问题用的到

板子

$v$是容量，$c[i]$是每个的体积，$w[i]$是每个的价值

for (int i = 1;i <= n;i++) {for (int j = v;j > c[i];j--) {dp[j] = max(dp[j], dp[j - c[i]] + w[i]);}} 

初始化条件

如果题目中要求，恰好装满背包，则$dp[0]=0$，$dp[i]=-\infty (1<=i<=v)$
若要求，只希望达到最大价值，不必须装满，则$dp[i]=0(0<=i<=v)$;

板子题1 农夫买草

洛谷p2925

ACcode

#include<bits/stdc++.h>using namespace std;const int M = 3e5 + 9;
int dp[M];int main()
{/* for (int i = 1;i <= n;i++) {for (int j = v;j >= 0;j--) {f[j] = max(f[j], f[j - c[i]] + w[i]);}}for (int i = 1;i <= n;i++) {for (int j = v;j >= c[i];j--) {f[j] = max(f[j], f[j - c[i]] + w[i]);}} */ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int a, b;cin >> a >> b;vector<int>c(b + 3);for (int i = 1;i <= b;i++)cin >> c[i];for (int i = 1;i <= b;i++) {for (int j = a;j >= c[i];j--) {dp[j] = max(dp[j], dp[j - c[i]]+c[i]);}if (dp[a] == a) {cout << a;return 0;}}cout << dp[a];return 0;
}

板子题2 采药

洛谷p1048

AC code

using namespace std;const int M = 1005;
int dp[M];int main()
{/* for (int i = 1;i <= n;i++) {for (int j = v;j > c[i];j--) {dp[j] = max(dp[j], dp[j - c[i]] + w[i]);}} */ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int a, b;cin >> a >> b;vector < pair<int, int> >c(b + 3);for (int i = 1;i <= b;i++)cin >> c[i].first >> c[i].second;for (int i = 1;i <= b;i++) {for (int j = a;j >= c[i].first;j--) {dp[j] = max(dp[j], dp[j - c[i].first] + c[i].second);}}cout << dp[a];return 0;
}