背包问题

01背包

一维费用

题目

朴素

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
#include <iostream>

using namespace std;

const int N = 1010;

int n, m, f[N][N], v[N], w[N];

int main()
{
    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 = 0; 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];

    return 0;
}

优化

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
#include <iostream>

using namespace std;

const int N = 1010;

int n, m, f[N], v[N], w[N];

int main()
{
    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], f[j - v[i]] + w[i]);

    cout << f[m];

    return 0;
}

二维费用

题目
朴素

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
#include <iostream>

using namespace std;

const int N = 1010, M = 110;

int f[N][M][M], v1[N], v2[N], w[N];
int n, m1, m2;

int main()
{
    cin >> n >> m1 >> m2;

    for (int i = 1; i <= n; i ++ ) cin >> v1[i] >> v2[i] >> w[i];

    for (int i = 1; i <= n; i ++ )
        for (int j = 0; j <= m1; j ++ )
            for (int k = 0; k <= m2; k ++ )
            {
                f[i][j][k] = f[i - 1][j][k];
                if (j >= v1[i] && k >= v2[i])
                    f[i][j][k] = max(f[i][j][k], f[i - 1][j - v1[i]][k - v2[i]] + w[i]);
            }

    cout << f[n][m1][m2];

    return 0;
}

优化

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
#include <iostream>

using namespace std;

const int N = 1010, M = 110;

int f[M][M], v1[N], v2[N], w[N];
int n, m1, m2;

int main()
{
    cin >> n >> m1 >> m2;

    for (int i = 1; i <= n; i ++ ) cin >> v1[i] >> v2[i] >> w[i];

    for (int i = 1; i <= n; i ++ )
        for (int j = m1; j >= v1[i]; j -- )
            for (int k = m2; k >= v2[i]; k -- )
                f[j][k] = max(f[j][k], f[j - v1[i]][k - v2[i]] + w[i]);

    cout << f[m1][m2];

    return 0;
}

分组背包

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
#include <iostream>

using namespace std;

const int N = 110;

int n, m, s[N], v[N][N], w[N][N], f[N][N];

int main()
{
    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 = 0; 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];

    return 0;
}

完全背包

题目
朴素

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
#include <iostream>

using namespace std;

const int N = 1010;

int n, m, f[N][N], v[N], w[N];

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

    cout << f[n][m];

    return 0;
}

优化

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
#include <iostream>

using namespace std;

const int N = 1010;

int n, m, f[N], v[N], w[N];

int main()
{
    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], f[j - v[i]] + w[i]);

    cout << f[m];

    return 0;
}

01背包求方案

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
#include <iostream>

using namespace std;

const int N = 1010;

//f[i, j]:从后i个物品中选, 总体积不超过j的选法的价值最大值
int f[N][N], v[N], w[N];
int n, m;

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

    for (int i = n; i >= 1; i -- )
    {
        for (int j = 0; 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]);
        }
    }

    for (int i = 1; i <= n; i ++ )
        if (m >= v[i] && f[i][m] == f[i + 1][m - v[i]] + w[i])
        {
            printf("%d ", i);
            m -= v[i];
        }

    return 0;
}

背包问题求方案数

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
#include <iostream>

using namespace std;

const int N = 1010, MOD = 1e9 + 7;

int f[N], g[N];
int v[N], w[N], n, m;

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

多重背包

时间复杂度(O(nms))

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
#include <iostream>

using namespace std;

const int N = 110;

int f[N], w[N], v[N], s[N], n, m;

int main()
{
    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 = m; j >= 0; j -- )
            for (int k = 0; k <= s[i]; k ++ )
                if (j >= k * v[i]) f[j] = max(f[j], f[j - k * v[i]] + k * w[i]);
                else break;

    cout << f[m];

    return 0;
}

时间复杂度(O(nmlogs))

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
#include <iostream>

using namespace std;

const int N = 12000;

int f[N], cnt, n, m, v[N], w[N];

//二进制转为01背包
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
    {
        int a, b, s;
        scanf("%d%d%d", &a, &b, &s);
        for (int k = 1; k <= s; k <<= 1)
        {
            cnt ++ , v[cnt] = a * k, w[cnt] = b * k;
            s -= k;
        }
        if (s) cnt ++, v[cnt] = a * s, w[cnt] = b * s;
    }

    for (int i = 1; i <= cnt; i ++ )
        for (int j = m; j >= v[i]; j -- )
            f[j] = max(f[j], f[j - v[i]]+ w[i]);

    cout << f[m];

    return 0;
}

单调队列优化(接近O(nm))

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <iostream>
#include <cstring>

using namespace std;

const int N = 20010;

int n, m;
int f[N], g[N], q[N];

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
    {
        int v, w, s;
        cin >> v >> w >> s;
        memcpy(g, f, sizeof f);

        for (int j = 0; j < v; j ++ )  //枚举余数
        {
            int hh = 0, tt = -1;
            for (int k = j; k <= m; k += v)
            {
                while (hh <= tt && q[hh] < k - s * v) hh ++ ;   //保持窗口大小
                while (hh <= tt && g[q[tt]] + (k - q[tt]) / v * w <= g[k]) tt -- ;   //保持单调
                q[ ++ tt] = k;
                f[k] = g[q[hh]] + (k - q[hh]) / v * w;
            }
        }
    }

    cout << f[m];

    return 0;
}

混合背包

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <iostream>

using namespace std;

const int N = 12010;

//f[i, j]:从前i个物品中选, 总体积不超过j的最大价值
int f[N], n, m;

int main()
{
    cin >> n >> m;

    for (int i = 1; i <= n; i ++ )
    {
        int v, w, s;
        scanf("%d%d%d", &v, &w, &s);

        if (s == -1) s = 1;  //01背包转多重背包
        if (s == 0) s = m / v;   //完全背包转多重背包

        for (int k = 1; k <= s; k *= 2)
        {
            for (int j = m; j >= k * v; j -- )
                f[j] = max(f[j], f[j - k * v] + k * w);
            s -= k;
        }
        if (s)
            for (int j = m; j >= s * v; j -- )
                f[j] = max(f[j], f[j - s * v] + s * w);
    }

    cout << f[m];

    return 0;
}

有依赖的背包问题

题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
#include <iostream>
#include <cstring>

using namespace std;

const int N = 110;

//f[i, j]:以i为根节点, 总体积不超过j的最大价值
int f[N][N];
int e[N], ne[N], h[N], idx;
int n, m, v[N], w[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void dfs(int u)
{
    for (int i = v[u]; i <= m; i ++ ) f[u][i] = w[u];  //选u

    //分组背包
    for (int i = h[u]; i != -1; i = ne[i])   //循环物品组
    {
        int son = e[i];
        dfs(son);

        for (int j = m; j >= v[u]; j -- )  //循环体积  <v[u]无法选子树
            for (int k = 0; k <= j - v[u]; k ++ )   //循环决策   >j - v[u]无法选u
                f[u][j] = max(f[u][j], f[u][j - k] + f[son][k]);
    }
}

int main()
{
    memset(h, -1, sizeof h);

    cin >> n >> m;
    int root;
    for (int i = 1; i <= n; i ++ )
    {
        int p;
        cin >> v[i] >> w[i] >> p;
        if (p == -1) root = i;
        else add(p, i);
    }

    dfs(root);

    cout << f[root][m];

    return 0;
}