01 背包和完全背包是重點，分組背包、二維費用是 01 背包的擴展，多重背包是受限制的完全背包

01 背包

解題思路

代碼

原始做法

#include <iostream>

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;
}

優化空間

#include <iostream>

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

解題思路

代碼

原始做法 O(N^3)

#include <iostream>

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;
}

優化時間 O(N^2)

#include <iostream>

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;
}

優化空間

#include <iostream>

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

解題思路

樸素的多重背包，在樸素的完全背包的基礎，加一個次數限制，即可

樸素做法：一個一個拆 O ( N ) O(N) O(N)

二進制優化：類似于快速冪的拆法 O ( l o g N ) O(logN) O(logN)，拆成多組

比如說：a 物品有 13 個

樸素做法就是一個一個的試，最終試到 13 個

二進制優化，則是拆成 5 組，{1, 2, 4, 8, 1} 嘗試，因為由 {1, 2, 4, 8, 1} 這個集合里面的子集能湊 0 ~ 13 的任意一個數

代碼

原始做法 O(N^3)

#include <iostream>

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;
}

二進制優化 O(N^2logN)

#include <iostream>

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;
}

在二進制優化上再優化空間

#include <iostream>

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;
}

分組背包

解題思路

分組背包就是擴展了的 01 背包，狀態轉移與 01 一致，只是拓展了分組內物品的列舉

代碼

原始做法

#include <iostream>

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;
}

優化空間

#include <iostream>

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

解題思路

01 背包與分組背包和完全背包混合，這三者的狀態表示是相同的，根據判斷背包的型別，分別做狀態轉移即可

代碼

原始做法 O(N^3)

#include <iostream>

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;
}

優化 O(N^2logN)

#include <iostream>

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;
}

二維費用背包

解題思路

代碼

原始做法

#include <iostream>
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;
}

優化空間

#include <iostream>
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;
}