刷透背包（01 背包，完全背包，多重背包，分组背包，混合背包，二维费用背包）

01 背包和完全背包是重点，分组背包、二维费用是 01 背包的扩展，多重背包是受限制的完全背包

01 背包

解题思路

代码

原始做法

#include 

using namespace std;

const int N = 1010;
int v[N], w[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            f[i][j] = f[i - 1][j];
            if (j >= v[i]) f[i][j] = max(f[i - 1][j - v[i]] + w[i], f[i][j]);
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

优化空间

#include 

using namespace std;

const int N = 1010;
int v[N], w[N], f[N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = m; j >= v[i]; j--) f[j] = max(f[j - v[i]] + w[i], f[j]);
    }
    cout << f[m] << endl;
    
    return 0;
}

完全背包

解题思路

代码

原始做法 O(N^3)

#include 

using namespace std;

const int N = 1010;
int v[N], w[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            for (int k = 0; j >= k * v[i]; k++) 
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

优化时间 O(N^2)

#include 

using namespace std;

const int N = 1010;
int v[N], w[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            f[i][j] = f[i - 1][j];
            if (j >= v[i]) f[i][j] = max(f[i - 1][j], f[i][j - v[i]] + w[i]);
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

优化空间

#include 

using namespace std;

const int N = 1010;
int v[N], w[N], f[N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = v[i]; j <= m; j++) {
     
            f[j] = max(f[j - v[i]] + w[i], f[j]);
        }
    }
    cout << f[m] << endl;
    
    return 0;
}

多重背包

解题思路

朴素的多重背包，在朴素的完全背包的基础，加一个次数限制，即可

朴素做法：一个一个拆 $O(N)$

二进制优化：类似于快速幂的拆法 $O(logN)$，拆成多组

比如说：a 物品有 13 个

朴素做法就是一个一个的试，最终试到 13 个

二进制优化，则是拆成 5 组，{1, 2, 4, 8, 1} 尝试。因为由 {1, 2, 4, 8, 1} 这个集合里面的子集能凑 0 ~ 13 的任意一个数

代码

原始做法 O(N^3)

#include 

using namespace std;

const int N = 110;
int v[N], w[N], s[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> s[i];

    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            for (int k = 0; j >= k * v[i] && k <= s[i]; k++) {
     
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);
            }
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

二进制优化 O(N^2logN)

#include 

using namespace std;

const int N = 10000;
int v[N], w[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
     
        int a, b, s;
        cin >> a >> b >> s;
        for (int k = 1; k <= s; s -= k, k <<= 1) {
     
            v[cnt] = a * k, w[cnt] = b * k;
            cnt++;
        }
        if (s) {
     
            v[cnt] = a * s, w[cnt] = b * s;
            cnt++;
        }
    }
    
    n = cnt;
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            f[i][j] = f[i - 1][j];
            if (j >= v[i]) f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]);
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

在二进制优化上再优化空间

#include 

using namespace std;

const int N = 10000;
int v[N], w[N], f[N];

int main() {
     
    int n, m;
    cin >> n >> m;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
     
        int a, b, s;
        cin >> a >> b >> s;
        for (int k = 1; k <= s; s -= k, k <<= 1) {
     
            v[cnt] = a * k, w[cnt] = b * k;
            cnt++;
        }
        if (s) {
     
            v[cnt] = a * s, w[cnt] = b * s;
            cnt++;
        }
    }
    
    n = cnt;
    for (int i = 1; i <= n; i++) {
     
        for (int j = m; j >= v[i]; j--) {
     
            f[j] = max(f[j], f[j - v[i]] + w[i]);
        }
    }
    cout << f[m] << endl;
    
    return 0;
}

分组背包

解题思路

分组背包就是扩展了的 01 背包，状态转移与 01 一致。只是拓展了分组内物品的枚举

代码

原始做法

#include 

using namespace std;

const int N = 110;
int v[N][N], w[N][N], s[N], f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
     
        cin >> s[i];
        for (int j = 1; j <= s[i]; j++) cin >> v[i][j] >> w[i][j];
    }
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= m; j++) {
     
            f[i][j] = f[i - 1][j];
            for (int k = 1; k <= s[i]; k++) {
     
                if (j >= v[i][k]) f[i][j] = max(f[i][j], f[i - 1][j - v[i][k]] + w[i][k]);
            }
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

优化空间

#include 

using namespace std;

const int N = 110;
int v[N][N], w[N][N], s[N], f[N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
     
        cin >> s[i];
        for (int j = 1; j <= s[i]; j++) cin >> v[i][j] >> w[i][j];
    }
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = m; j >= 0; j--) {
     
            for (int k = 1; k <= s[i]; k++) {
     
                if (j >= v[i][k]) f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
            }
        }
    }
    cout << f[m] << endl;
    
    return 0;
}

混合背包

解题思路

01 背包与分组背包和完全背包混合，这三者的状态表示是相同的，根据判断背包的类型，分别做状态转移即可

代码

原始做法 O(N^3)

#include 

using namespace std;

const int N = 1010;
int f[N][N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
     
        int v, w, s;
        cin >> v >> w >> s;
        if (s == -1) {
      // 01 背包
            for (int j = 1; j <= m; j++) {
     
                f[i][j] = f[i - 1][j];
                if (j >= v) f[i][j] = max(f[i][j], f[i - 1][j - v] + w);
            }
        } else if (s == 0) {
      // 完全背包
            for (int j = 1; j <= m; j++) {
     
                f[i][j] = f[i - 1][j];
                if (j >= v) f[i][j] = max(f[i - 1][j], f[i][j - v] + w);
            }
        } else {
      // 多重背包
            for (int j = 1; j <= m; j++) {
     
                for (int k = 0; j >= k * v && k <= s; k++) {
     
                    f[i][j] = max(f[i][j], f[i - 1][j - k * v] + k * w);
                }
            }
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

优化 O(N^2logN)

#include 

using namespace std;

const int N = 1010;
int f[N];

int main() {
     
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
     
        int v, w, s;
        cin >> v >> w >> s;
        if (s == -1) {
      // 01 背包
            for (int j = m; j >= v; j--) {
     
                if (j >= v) f[j] = max(f[j], f[j - v] + w);
            }
        } else if (s == 0) {
      // 完全背包
            for (int j = v; j <= m; j++) {
     
                f[j] = max(f[j], f[j - v] + w);
            }
        } else {
      // 多重背包
            for (int k = 1; k <= s; s -= k, k <<= 1) {
     
                for (int j = m; j >= k * v; j--) f[j] = max(f[j], f[j - k * v] + k * w);
            }
            if (s) {
     
                for (int j = m; j >= s * v; j--) f[j] = max(f[j], f[j - s * v] + s * w);
            }
        }
    }
    cout << f[m] << endl;
    
    return 0;
}

二维费用背包

解题思路

代码

原始做法

#include 
using namespace std;

const int N = 1010, M = 110;
int v[N], m[N], w[N];
int f[N][M][M];

int main() {
     
    int n, a, b;
    cin >> n >> a >> b;
    for (int i = 1; i <= n; i++) cin >> v[i] >> m[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = 1; j <= a; j++) {
     
            for (int k = 1; k <= b; k++) {
     
                f[i][j][k] = f[i - 1][j][k];
                if (j >= v[i] && k >= m[i])
                    f[i][j][k] = max(f[i][j][k], f[i - 1][j - v[i]][k - m[i]] + w[i]);
            }
        }
    }
    cout << f[n][a][b] << endl;

    return 0;
}

优化空间

#include 
using namespace std;

const int N = 1010, M = 110;
int v[N], m[N], w[N];
int f[M][M];

int main() {
     
    int n, a, b;
    cin >> n >> a >> b;
    for (int i = 1; i <= n; i++) cin >> v[i] >> m[i] >> w[i];
    
    for (int i = 1; i <= n; i++) {
     
        for (int j = a; j >= v[i]; j--) {
     
            for (int k = b; k >= m[i]; k--) {
     
                f[j][k] = max(f[j][k], f[j - v[i]][k - m[i]] + w[i]);
            }
        }
    }
    cout << f[a][b] << endl;

    return 0;
}