5.分组背包问题

Posted 2020-12-14 fx1998

 

 分组背包问题是物品有n组，每组里面有若干个物品，每组里最多选一个。

dp[i][j]只从前i组物品中选，且总体积不大于j的最大价值

状态计算：枚举第i组物品选哪个或不选

不选第i组物品，选第i组的第1个物品，选第i组的第2个物品，等等等等

 二维做法

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 110; //一共有100组物品 
 4 int s[N]; //每组物品所含的物品数 
 5 int v[N][N], w[N][N];
 6 int dp[N][N];
 7 int main() {
 8     int n, m;
 9     cin >> n >> m;
10     for (int i = 1; i <= n; i++) {
11         cin >> s[i];
12         for (int j = 1; j <= s[i]; j++) {
13             cin >> v[i][j] >> w[i][j];
14         }
15     }
16     for (int i = 1; i <= n; i++) {
17         for (int j = 0; j <= m; j++) {
18             dp[i][j] = dp[i - 1][j]; //不选 
19             for (int k = 1; k <= s[i]; k++) {
20                 if (j >= v[i][k]) {
21                     dp[i][j] = max(dp[i][j], dp[i - 1][j - v[i][k]] + w[i][k]);
22                 }
23             }
24         }
25     }
26     cout << dp[n][m] << endl;
27     return 0;
28 }

 

一维做法

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 110;
 4 int s[N], v[N][N], w[N][N], dp[N];
 5 int main() {
 6     int n, m;
 7     cin >> n >> m;
 8     for (int i = 1; i <= n; i++) {
 9         cin >> s[i];
10         for (int j = 1; j <= s[i]; j++) {
11             cin >> v[i][j] >> w[i][j];
12         }
13     }
14     for (int i = 1; i <= n; i++) {
15         for (int j = m; j >= 0; j--) {
16             for (int k = 1; k <= s[i]; k++) {
17                 if (j >= v[i][k]) {
18                     dp[j] = max(dp[j], dp[j - v[i][k]] + w[i][k]);
19                 }
20             }
21         }
22     }
23     cout << dp[m] << endl;
24     return 0;
25 }