特征

混合背包，就是把01，完全，多重背包混合起来
物品一共有三类：

第一类物品只能用1次（01背包）；
第二类物品可以用无限次（完全背包）；
第三类物品最多只能用 si
次（多重背包）；

板子

for (int i = 1;i <= n;i++) {cin >> c[i] >> w[i] >> s;if (s == 0) {//完全背包for (int j = c[i];j <= v;j++) {dp[j] = max(dp[j], dp[j - c[i]] + w[i]);}}else if (s == 1) {//01背包for (int j = v;j >= c[i];j--) {dp[j] = max(dp[j], dp[j - c[i]] + w[i]);}}else {//多重背包int num = min(s, v / c[i]);for (int k = 1;num > 0;k <<= 1) {if (k > num)k = num;num -= k;for (int j = v;j >= c[i] * k;j--) {dp[j] = max(dp[j], dp[j - c[i] * k] + w[i] * k);}}}}

板子题1

acwing混合背包

ACcode

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